9 ROTATION OF RIGID BODIES 9.1 9.2 IDENTIFY: s = rθ , with θ in radians SET UP: π rad = 180° s 1.50 m EXECUTE: (a) θ = = = 0.600 rad = 34.4° r 2.50 m s 14.0 cm (b) r = = = 6.27 cm θ (128°)(π rad/180°) (c) s = rθ = (1.50 m)(0.700 rad) = 1.05 m EVALUATE: An angle is the ratio of two lengths and is dimensionless But, when s = rθ is used, θ must be in radians Or, if θ = s /r is used to calculate θ , the calculation gives θ in radians IDENTIFY: θ − θ = ωt , since the angular velocity is constant SET UP: rpm = (2π /60) rad/s EXECUTE: (a) ω = (1900)(2π rad/60 s) = 199 rad/s (b) 35° = (35°)(π /180°) = 0.611 rad t = EVALUATE: In t = both θ − θ and ω 9.3 θ − θ 0.611 rad = = 3.1 × 10−3 s ω 199 rad/s θ − θ0 we must use the same angular measure (radians, degrees or revolutions) for ω dω z Writing Eq (2.16) in terms of angular quantities gives θ − θ0 = ∫ tt2 ω z dt dt n +1 d n t t = nt n − and ∫ t n dt = n +1 dt IDENTIFY α z (t ) = SET UP: EXECUTE: (a) A must have units of rad/s and B must have units of rad/s3 (b) α z (t ) = Bt = (3.00 rad/s3 )t (i) For t = 0, α z = (ii) For t = 5.00 s, α z = 15.0 rad/s (c) θ − θ1 = ∫ t2 ( A + Bt ) dt = A(t2 − t1 ) + 13 B(t23 − t13 ) For t1 = and t2 = 2.00 s, t θ − θ1 = (2.75 rad/s)(2.00 s) + 13 (1.50 rad/s3 )(2.00 s)3 = 9.50 rad EVALUATE: Both α z and ω z are positive and the angular speed is increasing 9.4 IDENTIFY: α z = dω z /dt α av-z = SET UP: Δω z Δt d (t ) = 2t dt EXECUTE: (a) α z (t ) = dω z = − β t = (− 1.60 rad/s3 )t dt (b) α z (3.0 s) = ( − 1.60 rad/s3 )(3.0 s) = − 4.80 rad/s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-1 9-2 Chapter 9.5 ω z (3.0 s) − ω z (0) − 2.20 rad/s − 5.00 rad/s = = − 2.40 rad/s , 3.0 s 3.0 s which is half as large (in magnitude) as the acceleration at t = 3.0 s α (0) + α z (3.0 s) EVALUATE: α z (t ) increases linearly with time, so α av- z = z α z (0) = IDENTIFY and SET UP: Use Eq (9.3) to calculate the angular velocity and Eq (9.2) to calculate the average angular velocity for the specified time interval EXECUTE: θ = γ t + β t ; γ = 0.400 rad/s, β = 0.0120 rad/s3 α av- z = dθ = γ + 3β t dt (b) At t = 0, ω z = γ = 0.400 rad/s (a) ω z = (c) At t = 5.00 s, ω z = 0.400 rad/s + 3(0.0120 rad/s3 )(5.00 s)2 = 1.30 rad/s ω av- z = Δθ θ − θ1 = t2 − t1 Δt For t1 = 0, θ1 = For t2 = 5.00 s, θ = (0.400 rad/s)(5.00 s) + (0.012 rad/s3 )(5.00 s)3 = 3.50 rad 3.50 rad − = 0.700 rad/s 5.00 s − EVALUATE: The average of the instantaneous angular velocities at the beginning and end of the time interval is 12 (0.400 rad/s + 1.30 rad/s) = 0.850 rad/s This is larger than ω av- z , because ω z (t ) is increasing So ω av- z = faster than linearly 9.6 ω z (t ) = IDENTIFY: dθ dω z Δθ α z (t ) = ωav − z = dt dt Δt SET UP: ω z = (250 rad/s) − (40.0 rad/s )t − (4.50 rad/s3 )t α z = − (40.0 rad/s ) − (9.00 rad/s3 )t (a) Setting ω z = results in a quadratic in t The only positive root is t = 4.23 s EXECUTE: (b) At t = 4.23 s, α z = − 78.1 rad/s (c) At t = 4.23 s, θ = 586 rad = 93.3 rev (d) At t = 0, ω z = 250 rad/s (e) ω av- z = 586 rad = 138 rad/s 4.23 s EVALUATE: Between t = and t = 4.23 s, ω z decreases from 250 rad/s to zero ω z is not linear in t, so 9.7 ω av-z is not midway between the values of ω z at the beginning and end of the interval dθ dω z IDENTIFY: ω z (t ) = α z (t ) = Use the values of θ and ω z at t = and α z at 1.50 s to calculate dt dt a, b, and c SET UP: d n t = nt n − dt EXECUTE: (a) ω z (t ) = b − 3ct α z (t ) = − 6ct At t = 0, θ = a = π /4 rad and ω z = b = 2.00 rad/s At t = 1.50 s, α z = − 6c(1.50 s) = 1.25 rad/s and c = − 0.139 rad/s3 (b) θ = π /4 rad and α z = at t = (c) α z = 3.50 rad/s at t = − θ= π αz 6c =− 3.50 rad/s 6( − 0.139 rad/s3 ) = 4.20 s At t = 4.20 s, rad + (2.00 rad/s)(4.20 s) − (− 0.139 rad/s3 )(4.20 s)3 = 19.5 rad © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9.8 9-3 ω z = 2.00 rad/s − 3(− 0.139 rad/s3 )(4.20 s) = 9.36 rad/s EVALUATE: θ , ω z and α z all increase as t increases dω z IDENTIFY: α z = θ − θ = ω av- z t When ω z is linear in t, ω av-z for the time interval t1 to t2 is dt ω av- z = ω z1 + ω z t2 − t1 SET UP: From the information given, ω z (t ) = −6.00 rad/s + (2.00 rad/s )t EXECUTE: (a) The angular acceleration is positive, since the angular velocity increases steadily from a negative value to a positive value (b) It takes 3.00 seconds for the wheel to stop (ω z = 0) During this time its speed is decreasing For the next 4.00 s its speed is increasing from rad/s to + 8.00 rad/s − 6.00 rad/s + 8.00 rad/s = 1.00 rad/s θ − θ = ω av-z t then leads to displacement of 7.00 rad after 7.00 s EVALUATE: When α z and ω z have the same sign, the angular speed is increasing; this is the case for (c) The average angular velocity is 9.9 t = 3.00 s to t = 7.00 s When α z and ω z have opposite signs, the angular speed is decreasing; this is the case between t = and t = 3.00 s IDENTIFY: Apply the constant angular acceleration equations SET UP: Let the direction the wheel is rotating be positive EXECUTE: (a) ω z = ω0 z + α z t = 1.50 rad/s + (0.300 rad/s2 )(2.50 s) = 2.25 rad/s (b) θ − θ = ω0 z t + 12 α z t = (1.50 rad/s)(2.50 s) + 12 (0.300 rad/s )(2.50 s) = 4.69 rad 9.10 ⎛ ω + ω z ⎞ ⎛ 1.50 rad/s + 2.25 rad/s ⎞ EVALUATE: θ − θ0 = ⎜ z ⎟t = ⎜ ⎟ (2.50 s) = 4.69 rad, the same as calculated 2 ⎝ ⎠ ⎝ ⎠ with another equation in part (b) IDENTIFY: Apply the constant angular acceleration equations to the motion of the fan (a) SET UP: ω0 z = (500 rev/min)(1 min/60 s) = 8.333 rev/s, ωz = (200 rev/min)(1 min/60 s) = 3.333 rev/s, t = 4.00 s, α z = ? ω z = ω0 z + α z t ω − ω0 z 3.333 rev/s − 8.333 rev/s = = − 1.25 rev/s EXECUTE: α z = z t 4.00 s θ − θ0 = ? θ − θ = ω0 z t + 12 α z t = (8.333 rev/s)(4.00 s) + 12 (− 1.25 rev/s )(4.00 s) = 23.3 rev (b) SET UP: ω z = (comes to rest); ω0 z = 3.333 rev/s; α z = − 1.25 rev/s ; t =? ω z = ω0 z + α z t EXECUTE: t = ω z − ω0 z − 3.333 rev/s = = 2.67 s αz − 1.25 rev/s EVALUATE: The angular acceleration is negative because the angular velocity is decreasing The average angular velocity during the 4.00 s time interval is 350 rev/min and θ − θ = ω av-z t gives θ − θ = 23.3 rev, which checks 9.11 IDENTIFY: Apply the constant angular acceleration equations to the motion The target variables are t and θ − θ SET UP: (a) α z = 1.50 rad/s ; ω0 z = (starts from rest); ω z = 36.0 rad/s; t = ? ω z = ω0 z + α z t © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-4 Chapter EXECUTE: t = ω z − ω0 z 36.0 rad/s − = = 24.0 s αz 1.50 rad/s (b) θ − θ0 = ? θ − θ = ω0 z t + 12 α z t = + 12 (1.50 rad/s )(24.0 s) = 432 rad θ − θ = 432 rad(1 rev/2π rad) = 68.8 rev EVALUATE: We could use θ − θ = 12 (ω z + ω0 z )t to calculate θ − θ0 = 12 (0 + 36.0 rad/s)(24.0 s) = 432 rad, 9.12 which checks IDENTIFY: In part (b) apply the equation derived in part (a) SET UP: Let the direction the propeller is rotating be positive ω − ω0 z EXECUTE: (a) Solving Eq (9.7) for t gives t = z Rewriting Eq (9.11) as θ − θ0 = t (ω0 z + 12 α z t ) αz and substituting for t gives ⎛ ω − ω0 z ⎞ ⎛ 1 ⎞ ⎛ ω + ω0 z ⎞ θ − θ0 = ⎜ z (ω z − ω0 z ) ⎜ z (ω z2 − ω02z ), ⎟ ⎜ ω0 z + (ω z − ω0 z ) ⎟ = ⎟= 2 ⎠ αz ⎝ ⎠ 2α z ⎝ αz ⎠⎝ which when rearranged gives Eq (9.12) ⎛ ⎞ ⎞ 2 2 1⎛ (b) α z = 12 ⎜ ⎟ ω z − ω0 z = ⎜ ⎟ (16.0 rad/s) − (12.0 rad/s) = 8.00 rad/s − θ θ 00 rad ⎝ ⎠ 0⎠ ⎝ ( ) ( ) ⎛ ω + ωz EVALUATE: We could also use θ − θ = ⎜ z ⎝ ⎞ ⎟ t to calculate t = 0.500 s Then ω z = ω0 z + α z t ⎠ gives α z = 8.00 rad/s , which agrees with our results in part (b) 9.13 IDENTIFY: Use a constant angular acceleration equation and solve for ω0 z SET UP: Let the direction of rotation of the flywheel be positive θ − θ0 60.0 rad EXECUTE: θ − θ0 = ω0 z t + 12 α z t gives ω0 z = − az t = − (2.25 rad/s2 )(4.00 s) = 10.5 rad/s t 4.00 s EVALUATE: At the end of the 4.00 s interval, ω z = ω0 z + α z t = 19.5 rad/s ⎛ ω0 z + ω z ⎞ ⎛ 10.5 rad/s + 19.5 rad/s ⎞ ⎟t = ⎜ ⎟ (4.00 s) = 60.0 rad, which checks 2 ⎝ ⎠ ⎝ ⎠ IDENTIFY: Apply the constant angular acceleration equations SET UP: Let the direction of the rotation of the blade be positive ω0 z = θ − θ0 = ⎜ 9.14 EXECUTE: ω z = ω0 z + α z t gives α z = ω z − ω0 z = 140 rad/s − = 23.3 rad/s 6.00 s t ⎛ ω0 z + ω z ⎞ ⎛ + 140 rad/s ⎞ (θ − θ0 ) = ⎜ ⎟t = ⎜ ⎟ (6.00 s) = 420 rad 2 ⎝ ⎠ ⎝ ⎠ EVALUATE: We could also use θ − θ = ω0 z t + 12 α z t This equation gives θ − θ0 = 12 (23.3 rad/s2 )(6.00 s)2 = 419 rad, in agreement with the result obtained above 9.15 IDENTIFY: Apply constant angular acceleration equations SET UP: Let the direction the flywheel is rotating be positive θ − θ = 200 rev, ω z = 500 rev/min = 8.333 rev/s, t = 30.0 s ⎛ ω + ωz ⎞ EXECUTE: (a) θ − θ = ⎜ z ⎟⎠ t gives ω z = 5.00 rev/s = 300 rpm ⎝ (b) Use the information in part (a) to find α z : ω z = ω0 z + α z t gives α z = − 0.1111 rev/s Then ω z = 0, ⎛ ω0 z + ω z ⎝ α z = − 0.1111 rev/s , ω z = 8.333 rev/s in ω z = ω0 z + α z t gives t = 75.0 s and θ − θ = ⎜ ⎞ ⎟t ⎠ gives θ − θ = 312 rev © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9.16 9-5 EVALUATE: The mass and diameter of the flywheel are not used in the calculation IDENTIFY: Apply the constant angular acceleration equations separately to the time intervals to 2.00 s and 2.00 s until the wheel stops (a) SET UP: Consider the motion from t = to t = 2.00 s: θ − θ0 = ?; ω0 z = 24.0 rad/s; α z = 30.0 rad/s ; t = 2.00 s EXECUTE: θ − θ = ω0 z t + 12 α z t = (24.0 rad/s)(2.00 s) + 12 (30.0 rad/s )(2.00 s) θ − θ = 48.0 rad + 60.0 rad = 108 rad Total angular displacement from t = until stops: 108 rad + 432 rad = 540 rad Note: At t = 2.00 s, ω z = ω0 z + α z t = 24.0 rad/s + (30.0 rad/s )(2.00 s) = 84.0 rad/s; angular speed when breaker trips (b) SET UP: Consider the motion from when the circuit breaker trips until the wheel stops For this calculation let t = when the breaker trips t = ?; θ − θ0 = 432 rad; ω z = 0; ω0 z = 84.0 rad/s (from part (a)) ⎛ ω0 z + ω z ⎞ ⎟t ⎝ ⎠ 2(θ − θ0 ) 2(432 rad) EXECUTE: t = = = 10.3 s ω0 z + ω z 84.0 rad/s + θ − θ0 = ⎜ The wheel stops 10.3 s after the breaker trips so 2.00 s + 10.3 s = 12.3 s from the beginning (c) SET UP: α z = ?; consider the same motion as in part (b): ω z = ω0 z + α z t ω z − ω0 z − 84.0 rad/s = = − 8.16 rad/s t 10.3 s EVALUATE: The angular acceleration is positive while the wheel is speeding up and negative while it is slowing down We could also use ω z2 = ω02z + 2α z (θ − θ ) to calculate EXECUTE: α z = αz = 9.17 ω z2 − ω02z − (84.0 rad/s) = = −8.16 rad/s for the acceleration after the breaker trips 2(θ − θ ) 2(432 rad) IDENTIFY: Apply Eq (9.12) to relate ω z to θ − θ0 SET UP: Establish a proportionality EXECUTE: From Eq (9.12), with ω0 z = 0, the number of revolutions is proportional to the square of the 9.18 initial angular velocity, so tripling the initial angular velocity increases the number of revolutions by 9, to 9.00 rev EVALUATE: We don’t have enough information to calculate α z ; all we need to know is that it is constant IDENTIFY: The linear distance the elevator travels, its speed and the magnitude of its acceleration are equal to the tangential displacement, speed and acceleration of a point on the rim of the disk s = rθ , v = rω and a = rα In these equations the angular quantities must be in radians SET UP: rev = 2π rad rpm = 0.1047 rad/s π rad = 180° For the disk, r = 1.25 m EXECUTE: (a) v = 0.250 m/s so ω = v 0.250 m/s = = 0.200 rad/s = 1.91 rpm r 1.25 m a 1.225 m/s = = 0.980 rad/s r 1.25 m s 3.25 m (c) s = 3.25 m θ = = = 2.60 rad = 149° r 1.25 m EVALUATE: When we use s = rθ , v = rω and atan = rα to solve for θ , ω and α , the results are in rad, (b) a = 18 g = 1.225 m/s α = rad/s and rad/s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-6 9.19 Chapter IDENTIFY: When the angular speed is constant, ω = θ /t vtan = rω , atan = rα and arad = rω In these equations radians must be used for the angular quantities SET UP: The radius of the earth is RE = 6.38 × 106 m and the earth rotates once in day = 86,400 s The orbit radius of the earth is 1.50 × 1011 m and the earth completes one orbit in y = 3.156 × 107 s When ω is constant, ω = θ /t 2π rad EXECUTE: (a) θ = rev = 2π rad in t = 3.156 × 107 s ω = = 1.99 × 10−7 rad/s 3.156 × 107 s 2π rad (b) θ = rev = 2π rad in t = 86,400 s ω = = 7.27 × 10−5 rad/s 86,400 s (c) v = rω = (1.50 × 1011 m)(1.99 × 10−7 rad/s) = 2.98 × 104 m/s (d) v = rω = (6.38 × 106 m)(7.27 × 10−5 rad/s) = 464 m/s (e) arad = rω = (6.38 × 106 m)(7.27 × 10−5 rad/s)2 = 0.0337 m/s atan = rα = α = since the angular 9.20 9.21 velocity is constant EVALUATE: The tangential speeds associated with these motions are large even though the angular speeds are very small, because the radius for the circular path in each case is quite large IDENTIFY: Linear and angular velocities are related by v = rω Use ω z = ω0 z + α z t to calculate α z SET UP: ω = v/r gives ω in rad/s 1.25 m/s 1.25 m/s = 50.0 rad/s, EXECUTE: (a) = 21.6 rad/s −3 25.0 × 10 m 58.0 × 10−3 m (b) (1.25 m/s)(74.0 min)(60 s/min ) = 5.55 km 21.55 rad/s − 50.0 rad/s = −6.41 × 10−3 rad/s (c) α z = (74.0 min)(60 s/min) EVALUATE: The width of the tracks is very small, so the total track length on the disc is huge IDENTIFY: Use constant acceleration equations to calculate the angular velocity at the end of two revolutions v = rω SET UP: rev = 4π rad r = 0.200 m EXECUTE: (a) ω z2 = ω02z + 2α z (θ − θ ) ω z = 2α z (θ − θ ) = 2(3.00 rad/s )(4π rad) = 8.68 rad/s arad = rω = (0.200 m)(8.68 rad/s)2 = 15.1 m/s (b) v = rω = (0.200 m)(8.68 rad/s) = 1.74 m/s arad = v (1.74 m/s) = = 15.1 m/s 0.200 m r EVALUATE: rω and v 2/r are completely equivalent expressions for arad 9.22 IDENTIFY: v = rω and atan = rα SET UP: The linear acceleration of the bucket equals atan for a point on the rim of the axle ⎛ 7.5 rev ⎞⎛ ⎞⎛ 2π rad ⎞ EXECUTE: (a) v = Rω 2.00 cm/s = R ⎜ ⎟⎜ ⎟⎜ ⎟ gives R = 2.55 cm ⎝ ⎠⎝ 60 s ⎠⎝ rev ⎠ D = R = 5.09 cm atan 0.400 m/s = = 15.7 rad/s 0.0255 m R EVALUATE: In v = Rω and atan = Rα , ω and α must be in radians IDENTIFY and SET UP: Use constant acceleration equations to find ω and α after each displacement Then use Eqs (9.14) and (9.15) to find the components of the linear acceleration EXECUTE: (a) at the start t = (b) atan = Rα α = 9.23 flywheel starts from rest so ω = ω0 z = atan = rα = (0.300 m)(0.600 rad/s ) = 0.180 m/s2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9-7 arad = rω = 2 a = arad + atan = 0.180 m/s (b) θ − θ0 = 60° atan = rα = 0.180 m/s Calculate ω : θ − θ = 60°(π rad/180°) = 1.047 rad; ω0 z = 0; α z = 0.600 rad/s ; ω z = ? ω z2 = ω02z + 2α z (θ − θ0 ) ω z = 2α z (θ − θ ) = 2(0.600 rad/s )(1.047 rad) = 1.121 rad/s and ω = ω z Then arad = rω = (0.300 m)(1.121 rad/s) = 0.377 m/s 2 a = arad + atan = (0.377 m/s ) + (0.180 m/s )2 = 0.418 m/s (c) θ − θ = 120° atan = rα = 0.180 m/s Calculate ω : θ − θ = 120°(π rad/180°) = 2.094 rad; ω0 z = 0; α z = 0.600 rad/s ; ω z = ? ω z2 = ω02z + 2α z (θ − θ0 ) ω z = 2α z (θ − θ0 ) = 2(0.600 rad/s )(2.094 rad) = 1.585 rad/s and ω = ω z Then arad = rω = (0.300 m)(1.585 rad/s) = 0.754 m/s 2 a = arad + atan = (0.754 m/s )2 + (0.180 m/s ) = 0.775 m/s 9.24 EVALUATE: α is constant so α tan is constant ω increases so arad increases IDENTIFY: Apply constant angular acceleration equations v = rω A point on the rim has both tangential and radial components of acceleration SET UP: atan = rα and arad = rω EXECUTE: (a) ω z = ω0 z + α z t = 0.250 rev/s + (0.900 rev/s )(0.200 s) = 0.430 rev/s (Note that since ω0 z and α z are given in terms of revolutions, it’s not necessary to convert to radians) (b) ω av- z Δt = (0.340 rev/s)(0.2 s) = 0.068 rev (c) Here, the conversion to radians must be made to use Eq (9.13), and ⎛ 0.750 m ⎞ v = rω = ⎜ ⎟ (0.430 rev/s)(2π rad/rev) = 1.01 m/s ⎝ ⎠ (d) Combining Eqs (9.14) and (9.15), 2 a = arad + atan = (ω 2r ) + (α r ) 2 a = ⎡⎣((0.430 rev/s)(2π rad/rev))2 (0.375 m) ⎤⎦ + ⎡⎣ (0.900 rev/s )(2π rad/rev)(0.375 m) ⎤⎦ a = 3.46 m/s EVALUATE: If the angular acceleration is constant, atan is constant but arad increases as ω increases 9.25 IDENTIFY: Use Eq (9.15) and solve for r SET UP: arad = rω so r = arad /ω , where ω must be in rad/s EXECUTE: arad = 3000 g = 3000(9.80 m/s ) = 29,400 m/s2 ⎛ ⎞⎛ 2π rad ⎞ ⎟⎜ ⎟ = 523.6 rad/s ⎝ 60 s ⎠⎝ rev ⎠ ω = (5000 rev/min) ⎜ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-8 Chapter Then r = 9.26 arad 29,400 m/s = 0.107 m (523.6 rad/s) EVALUATE: The diameter is then 0.214 m, which is larger than 0.127 m, so the claim is not realistic IDENTIFY: In part (b) apply the result derived in part (a) SET UP: arad = rω and v = rω ; combine to eliminate r ω2 = ⎛v⎞ EXECUTE: (a) arad = ω 2r = ω ⎜ ⎟ = ωv ⎝ω ⎠ (b) From the result of part (a), ω = arad 0.500 m/s = = 0.250 rad/s 2.00 m/s v EVALUATE: arad = rω and v = rω both require that ω be in rad/s, so in arad = ωv, ω is in rad/s 9.27 IDENTIFY: v = rω and arad = rω = v /r SET UP: 2π rad = rev, so π rad/s = 30 rev/min EXECUTE: (a) ω r = (1250 rev/min) rad/s ⎛⎜ 12.7 × 10 )⎜⎝ ( 30πrev/min −3 m⎞ ⎟⎟ = 0.831 m/s ⎠ v2 (0.831 m/s) = = 109 m/s r (12.7 × 10−3 m)/2 EVALUATE: In v = rω , ω must be in rad/s (b) 9.28 IDENTIFY: atan = rα , v = rω and arad = v /r θ − θ = ω av- z t SET UP: When α z is constant, ω av- z = ω0 z + ω z Let the direction the wheel is rotating be positive atan −10.0 m/s = = −50.0 rad/s 0.200 m r v (b) At t = 3.00 s, v = 50.0 m/s and ω = = 50.0 m/s = 250 rad/s and at t = 0, r 0.200 m EXECUTE: (a) α = v = 50.0 m/s + ( −10.0 m/s )(0 − 3.00 s) = 80.0 m/s, ω = 400 rad/s (c) ω av- z t = (325 rad/s)(3.00 s) = 975 rad = 155 rev (d) v = arad r = (9.80 m/s )(0.200 m) = 1.40 m/s This speed will be reached at time 50.0 m/s − 1.40 m/s = 4.86 s after t = 3.00 s, or at t = 7.86 s (There are many equivalent ways to this 10.0 m/s calculation.) EVALUATE: At t = 0, arad = rω = 3.20 × 104 m/s At t = 3.00 s, arad = 1.25 × 104 m/s For arad = g 9.29 the wheel must be rotating more slowly than at 3.00 s so it occurs some time after 3.00 s G G IDENTIFY and SET UP: Use Eq (9.15) to relate ω to arad and ∑ F = ma to relate arad to Frad Use Eq (9.13) to relate ω and v, where v is the tangential speed EXECUTE: (a) arad = rω and Frad = marad = mrω 2 ⎛ ω ⎞ ⎛ 640 rev/min ⎞2 =⎜ ⎟ =⎜ ⎟ = 2.29 Frad,1 ⎝ ω1 ⎠ ⎝ 423 rev/min ⎠ (b) v = rω v2 ω2 640 rev/min = = = 1.51 v1 ω1 423 rev/min (c) v = rω ⎛ ⎞⎛ 2π rad ⎞ ω = (640 rev/min) ⎜ ⎟⎜ ⎟ = 67.0 rad/s ⎝ 60 s ⎠⎝ rev ⎠ Frad,2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9-9 Then v = rω = (0.235 m)(67.0 rad/s) = 15.7 m/s arad = rω = (0.235 m)(67.0 rad/s) = 1060 m/s arad 1060 m/s = = 108; a = 108 g g 9.80 m/s EVALUATE: In parts (a) and (b), since a ratio is used the units cancel and there is no need to convert ω to rad/s In part (c), v and arad are calculated from ω , and ω must be in rad/s 9.30 IDENTIFY and SET UP: Use Eq (9.16) Treat the spheres as point masses and ignore I of the light rods EXECUTE: The object is shown in Figure 9.30a (a) r = (0.200 m) + (0.200 m) = 0.2828 m I = ∑ mi ri2 = 4(0.200 kg)(0.2828 m) I = 0.0640 kg ⋅ m Figure 9.30a (b) The object is shown in Figure 9.30b r = 0.200 m I = ∑ mi ri2 = 4(0.200 kg)(0.200 m) I = 0.0320 kg ⋅ m Figure 9.30b (c) The object is shown in Figure 9.30c r = 0.2828 m I = ∑ mi ri2 = 2(0.200 kg)(0.2828 m) I = 0.0320 kg ⋅ m Figure 9.30c 9.31 EVALUATE: In general I depends on the axis and our answer for part (a) is larger than for parts (b) and (c) It just happens that I is the same in parts (b) and (c) IDENTIFY: Use Table 9.2 The correct expression to use in each case depends on the shape of the object and the location of the axis SET UP: In each case express the mass in kg and the length in m, so the moment of inertia will be in kg ⋅ m EXECUTE: (a) (i) I = 13 ML2 = 13 (2.50 kg)(0.750 m) = 0.469 kg ⋅ m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-10 Chapter ML2 = (0.469 kg ⋅ m ) = 0.117 kg ⋅ m (iii) For a very thin rod, all of the mass is at the axis (ii) I = 12 and I = (b) (i) I = 52 MR = 25 (3.00 kg)(0.190 m)2 = 0.0433 kg ⋅ m (ii) I = 23 MR = 53 (0.0433 kg ⋅ m ) = 0.0722 kg ⋅ m (c) (i) I = MR = (8.00 kg)(0.0600 m) = 0.0288 kg ⋅ m (ii) I = 12 MR = 12 (8.00 kg)(0.0600 m) = 0.0144 kg ⋅ m 9.32 EVALUATE: I depends on how the mass of the object is distributed relative to the axis IDENTIFY: Treat each block as a point mass, so for each block I = mr 2, where r is the distance of the block from the axis The total I for the object is the sum of the I for each of its pieces SET UP: In part (a) two blocks are a distance L/2 from the axis and the third block is on the axis In part (b) two blocks are a distance L /4 from the axis and one is a distance 3L /4 from the axis EXECUTE: (a) I = 2m( L /2) = 12 mL2 11 mL2 (2 + 9) = mL2 16 16 EVALUATE: For the same object I is in general different for different axes IDENTIFY: I for the object is the sum of the values of I for each part SET UP: For the bar, for an axis perpendicular to the bar, use the appropriate expression from Table 9.2 For a point mass, I = mr , where r is the distance of the mass from the axis (b) I = 2m( L/4 ) + m(3L/4) = 9.33 EXECUTE: (a) I = I bar + I balls = ⎛L⎞ M bar L2 + 2mballs ⎜ ⎟ 12 ⎝2⎠ (4.00 kg)(2.00 m) + 2(0.500 kg)(1.00 m) = 2.33 kg ⋅ m 12 1 (b) I = mbar L2 + mball L2 = (4.00 kg)(2.00 m) + (0.500 kg)(2.00 m)2 = 7.33 kg ⋅ m 3 (c) I = because all masses are on the axis (d) All the mass is a distance d = 0.500 m from the axis and I= I = mbar d + 2mball d = M Total d = (5.00 kg)(0.500 m) = 1.25 kg ⋅ m 9.34 EVALUATE: I for an object depends on the location and direction of the axis IDENTIFY: Compare this object to a uniform disk of radius R and mass 2M SET UP: With an axis perpendicular to the round face of the object at its center, I for a uniform disk is the same as for a solid cylinder EXECUTE: (a) The total I for a disk of mass M and radius R, I = 12 (2 M ) R = MR Each half of the disk has the same I, so for the half-disk, I = 12 MR (b) The same mass M is distributed the same way as a function of distance from the axis (c) The same method as in part (a) says that I for a quarter-disk of radius R and mass M is half that of a half-disk of radius R and mass 2M , so I = 12 ( 12 [2 M ]R ) = 12 MR 9.35 EVALUATE: I depends on how the mass of the object is distributed relative to the axis, and this is the same for any segment of a disk IDENTIFY and SET UP: I = ∑ mi ri2 implies I = I rim + I spokes EXECUTE: I rim = MR = (1.40 kg)(0.300 m)2 = 0.126 kg ⋅ m Each spoke can be treated as a slender rod with the axis through one end, so I spokes = ( ML2 )= (0.280 kg)(0.300 m) = 0.0672 kg ⋅ m I = I rim + I spokes = 0.126 kg ⋅ m + 0.0672 kg ⋅ m = 0.193 kg ⋅ m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-18 Chapter ω z = 2α z (θ − θ ) = 2(18.85 rad/s )(973.9 rad) = 191.6 rad/s The horizontal velocity of the piece is v = rω = (0.120 m)(191.6 rad/s) = 23.0 m/s Now consider the projectile motion of the piece Take +y downward and use the vertical motion to find t Solving y − y0 = v0 yt + a yt for t gives t= 9.64 2( y − y0 ) 2(0.820 m) = = 0.4091 s Then x − x0 = v0 xt + a xt = (23.0 m/s)(0.4091 s) = 9.41 m 2 ay 9.8 m/s EVALUATE: Once the piece is free of the blade, the only force acting on it is gravity so its acceleration is g downward IDENTIFY and SET UP: Use Eqs (9.3) and (9.5) As long as α z > 0, ω z increases At the t when α z = 0, ω z is at its maximum positive value and then starts to decrease when α z becomes negative θ (t ) = γ t − β t ; γ = 3.20 rad/s , β = 0.500 rad/s3 (a) ω z (t ) = EXECUTE: dθ d (γ t − β t ) = = 2γ t − 3β t dt dt d ω z d (2γ t − 3β t ) = = 2γ − β t dt dt (c) The maximum angular velocity occurs when α z = (b) α z (t ) = 2γ − β t = implies t = 2γ γ 3.20 rad/s = = = 2.133 s β 3β 3(0.500 rad/s3 ) At this t, ω z = 2γ t − 3β t = 2(3.20 rad/s )(2.133 s) − 3(0.500 rad/s3 )(2.133 s) = 6.83 rad/s 9.65 The maximum positive angular velocity is 6.83 rad/s and it occurs at 2.13 s EVALUATE: For large t both ω z and α z are negative and ω z increases in magnitude In fact, ω z → −∞ at t → ∞ So the answer in (c) is not the largest angular speed, just the largest positive angular velocity IDENTIFY: The angular acceleration α of the disk is related to the linear acceleration a of the ball by t t 0 a = Rα Since the acceleration is not constant, use ω z − ω0 z = ∫ α z dt and θ − θ0 = ∫ ω z dt to relate θ , ω z , α z and t for the disk ω0 z = SET UP: ∫t n dt = n +1 t In a = Rα , α is in rad/s n +1 EXECUTE: (a) A = (b) α = a 1.80 m/s = = 0.600 m/s3 t 3.00 s a (0.600 m/s3 )t = = (2.40 rad/s3 )t R 0.250 m t 15.0 rad/s 1.20 rad/s3 (c) ω z = ∫ (2.40 rad/s3 )tdt = (1.20 rad/s3 )t ω z = 15.0 rad/s for t = t t 0 = 3.54 s (d) θ − θ0 = ∫ ω z dt = ∫ (1.20 rad/s3 )t dt = (0.400 rad/s3 )t For t = 3.54 s, θ − θ = 17.7 rad 9.66 EVALUATE: If the disk had turned at a constant angular velocity of 15.0 rad/s for 3.54 s it would have turned through an angle of 53.1 rad in 3.54 s It actually turns through less than half this because the angular velocity is increasing in time and is less than 15.0 rad/s at all but the end of the interval IDENTIFY and SET UP: The translational kinetic energy is K = 12 mv and the kinetic energy of the rotating flywheel is K = 12 I ω Use the scale speed to calculate the actual speed v From that calculate K for the car and then solve for ω that gives this K for the flywheel vtoy Ltoy EXECUTE: (a) = vscale Lreal © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9-19 ⎛ Ltoy ⎞ ⎛ 0.150 m ⎞ vtoy = vscale ⎜ ⎟ = (700 km/h) ⎜ ⎟ = 35.0 km/h L ⎝ 3.0 m ⎠ ⎝ real ⎠ vtoy = ( 35.0 km/h )(1000 m/1 km)(1 h/3600 s) = 9.72 m/s (b) K = 12 mv = 12 (0.180 kg)(9.72 m/s) = 8.50 J (c) K = 12 I ω gives that ω = EVALUATE: 9.67 2K 2(8.50 J) = = 652 rad/s I 4.00 × 10−5 kg ⋅ m K = 12 I ω gives ω in rad/s 652 rad/s = 6200 rev/min so the rotation rate of the flywheel is very large G G IDENTIFY: atan = rα , arad = rω Apply the constant acceleration equations and ∑ F = ma G 2 + atan SET UP: atan and arad are perpendicular components of a , so a = arad EXECUTE: (a) α = atan 2.00 m/s = = 0.0333 rad/s r 60.0 m (b) α t = (0.0333 rad/s )(6.00 s) = 0.200 rad/s (c) arad = ω 2r = (0.200 rad/s)2 (60.0 m) = 2.40 m/s (d) The sketch is given in Figure 9.67 2 (e) a = arad + atan = (2.40 m/s ) + (2.00 m/s ) = 3.12 m/s , and the magnitude of the force is F = ma = (1240 kg)(3.12 m/s ) = 3.87 kN ⎛a ⎞ ⎛ 2.40 ⎞ D (f) arctan ⎜ rad ⎟ = arctan ⎜ ⎟ = 50.2 a ⎝ 2.00 ⎠ ⎝ tan ⎠ G G EVALUATE: atan is constant and arad increases as ω increases At t = , a is parallel to v As t G G G increases, a moves toward the radial direction and the angle between a and v increases toward 90° Figure 9.67 9.68 IDENTIFY: Apply conservation of energy to the system of drum plus falling mass, and compare the results for earth and for Mars SET UP: K drum = 12 I ω K mass = 12 mv v = Rω so if K drum is the same, ω is the same and v is the same on both planets Therefore, K mass is the same Let y = at the initial height of the mass and take + y upward Configuration is when the mass is at its initial position and is when the mass has descended 5.00 m, so y1 = and y2 = - h, where h is the height the mass descends © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-20 Chapter EXECUTE: (a) K1 + U1 = K + U gives = K drum + K mass − mgh K drum + K mass are the same on both ⎛ 9.80 m/s ⎞ ⎛g ⎞ planets, so mg E hE = mg M hM hM = hE ⎜ E ⎟ = (5.00 m) ⎜ = 13.2 m ⎟ ⎜ ⎟ ⎝ gM ⎠ ⎝ 3.71 m/s ⎠ (b) mg M hM = K drum + K mass v = g M hM − 9.69 mv 2 = mg M hM − K drum and K drum 2(250.0 J) = 2(3.71 m/s )(13.2 m) − = 8.04 m/s m 15.0 kg EVALUATE: We did the calculations without knowing the moment of inertia I of the drum, or the mass and radius of the drum IDENTIFY and SET UP: All points on the belt move with the same speed Since the belt doesn’t slip, the speed of the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are related to the angular speed of each circular object by v = rω EXECUTE: Figure 9.69 (a) v1 = r1ω1 ω1 = (60.0 rev/s)(2π rad/1 rev) = 377 rad/s v1 = r1ω1 = (0.45 × 10-2 m)(377 rad/s) = 1.70 m/s (b) v1 = v2 r1ω1 = r2ω2 ω2 = ( r1/r2 )ω1 = (0.45 cm/1.80 cm)(377 rad/s) = 94.2 rad/s 9.70 EVALUATE: The wheel has a larger radius than the shaft so turns slower to have the same tangential speed for points on the rim IDENTIFY: The speed of all points on the belt is the same, so r1ω1 = r2ω2 applies to the two pulleys SET UP: The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is the angular velocity of the saw blade π rad/s = 30 rev/min ⎛ π rad/s ⎞⎛ 0.208 m ⎞ EXECUTE: (a) v2 = (2(3450 rev/min)) ⎜ ⎟⎜ ⎟ = 75.1 m/s ⎝ 30 rev/min ⎠⎝ ⎠ ⎛ ⎛ π rad/s ⎞ ⎞ ⎛ 0.208 m ⎞ (b) arad = ω 2r = ⎜ 2(3450 rev/min) ⎜ ⎟⎟ ⎜ ⎟ = 5.43 × 10 m/s , ⎝ 30 rev/min ⎠ ⎠ ⎝ ⎠ ⎝ so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity 9.71 EVALUATE: In v = rω and arad = rω , ω must be in rad/s IDENTIFY: Apply v = rω SET UP: Points on the chain all move at the same speed, so rrωr = rf ωf vr 5.00 m s = = 15.15 rad s r 0.330 m The angular velocity of the front wheel is ωf = 0.600 rev s = 3.77 rad s rr = rf (ωf /ωr ) = 2.99 cm EXECUTE: The angular velocity of the rear wheel is ωr = EVALUATE: The rear sprocket and wheel have the same angular velocity and the front sprocket and wheel have the same angular velocity rω is the same for both, so the rear sprocket has a smaller radius since it has a larger angular velocity The speed of a point on the chain is v = rrω r = (2.99 × 10−2 m)(15.15 rad/s) = 0.453 m/s The linear speed of the bicycle is 5.00 m/s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9.72 9-21 IDENTIFY: Use the constant angular acceleration equations, applied to the first revolution and to the first two revolutions SET UP: Let the direction the disk is rotating be positive rev = 2π rad Let t be the time for the first revolution The time for the first two revolutions is t + 0.750 s EXECUTE: (a) θ − θ = ω0 z t + 12 α z t applied to the first revolution and then to the first two revolutions gives 2π rad = 12 α zt and 4π rad = 12 α z (t + 0.750 s) Eliminating α z between these equations gives 2π rad (t + 0.750 s) 2t = (t + 0.750 s) 2t = ±(t + 0.750 s) The positive root is t2 0.750 s t= = 1.81 s −1 (b) 2π rad = 12 α zt and t = 1.81 s gives α z = 3.84 rad/s 4π rad = EVALUATE: At the start of the second revolution, ω0 z = (3.84 rad/s2 )(1.81 s) = 6.95 rad/s The distance 9.73 the disk rotates in the next 0.750 s is θ − θ = ω0 zt + 12 α zt = (6.95 rad/s)(0.750 s) + 12 (3.84 rad/s )(0.750 s) = 6.29 rad, which is two revolutions IDENTIFY and SET UP: Use Eq (9.15) to relate arad to ω and then use a constant acceleration equation to replace ω EXECUTE: (a) arad = rω , arad,1 = rω12 , arad,2 = rω22 Δarad = arad,2 − arad,1 = r (ω22 − ω12 ) One of the constant acceleration equations can be written ω22z = ω12z + 2α (θ − θ1 ), or ω22z − ω12z = 2α z (θ − θ1) Thus Δarad = r 2α z (θ − θ1 ) = 2rα z (θ − θ1), as was to be shown (b) α z = Δarad 85.0 m/s − 25.0 m/s = = 6.00 rad/s Then 2r (θ − θ1 ) 2(0.250 m)(20.0 rad) atan = rα = (0.250 m)(6.00 rad/s ) = 1.50 m/s EVALUATE: ω is proportional to α z and (θ − θ ) so arad is also proportional to these quantities arad increases while r stays fixed, ω z increases, and α z is positive IDENTIFY and SET UP: Use Eq (9.17) to relate K and ω and then use a constant acceleration equation to replace ω EXECUTE: (c) K = 12 I ω ; K = 12 I ω22 , K1 = 12 I ω12 ΔK = K − K1 = 12 I (ω22 − ω12 ) = 12 I (2α z (θ − θ1 )) = Iα z (θ − θ1 ), as was to be shown (d) I = ΔK 45.0 J − 20.0 J = = 0.208 kg ⋅ m α z (θ − θ1 ) (6.00 rad/s )(20.0 rad) EVALUATE: α z is positive, ω increases and K increases 9.74 IDENTIFY: I = I wood + I lead m = ρV , where ρ is the volume density and m = σ A, where σ is the area density SET UP: For a solid sphere, I = 52 mR For the hollow sphere (foil), I = 23 mR For a sphere, V = 43 π R3 and A = 4π R mw = ρ wVw = ρ w π R3 mL = σ L AL = σ L 4π R 2 2⎛ ⎞ ⎛ρ R ⎞ EXECUTE: I = mw R + mL R = ⎜ ρ w π R3 ⎟ R + (σ L 4π R ) R = π R ⎜ w + σ L ⎟ 5⎝ 3 ⎠ ⎝ ⎠ I= ⎡ (800 kg/m3 )(0.30 m) ⎤ 8π + 20 kg/m ⎥ = 4.61 kg ⋅ m (0.30 m)4 ⎢ ⎣⎢ ⎦⎥ EVALUATE: mW = 90.5 kg and I W = 3.26 kg ⋅ m mL = 22.6 kg and I L = 1.36 kg ⋅ m Even though the foil is only 20% of the total mass, its contribution to I is about 30% of the total © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-22 9.75 Chapter IDENTIFY: K = 12 I ω arad = rω m = ρV SET UP: For a disk with the axis at the center, I = 12 mR V = tπ R , where t = 0.100 m is the thickness of the flywheel ρ = 7800 kg/m3 is the density of the iron EXECUTE: (a) ω = 90.0 rpm = 9.425 rad/s I = 2K ω = 2(10.0 × 106 J) (9.425 rad/s)2 = 2.252 × 105 kg ⋅ m 1 1/4 m = ρV = ρπ R 2t I = mR = ρπ tR This gives R = ( I/ρπ t ) = 3.68 m and the diameter is 7.36 m 2 (b) arad = Rω = 327 m/s EVALUATE: In K = 12 I ω , ω must be in rad/s arad is about 33g; the flywheel material must have large 9.76 cohesive strength to prevent the flywheel from flying apart IDENTIFY: K = 12 I ω To have the same K for any ω the two parts must have the same I Use Table 9.2 for I SET UP: For a solid sphere, I solid = 25 M solid R For a hollow sphere, I hollow = 23 M hollow R EXECUTE: I solid = I hollow gives 9.77 M R2 solid = 23 M hollow R and M hollow = 35 M solid = 35 M EVALUATE: The hollow sphere has less mass since all its mass is distributed farther from the rotation axis 2π rad IDENTIFY: K = 12 I ω ω = , where T is the period of the motion For the earth’s orbital motion it T can be treated as a point mass and I = MR SET UP: The earth’s rotational period is 24 h = 86,164 s Its orbital period is yr = 3.156 × 107 s M = 5.97 × 1024 kg R = 6.38 × 106 m EXECUTE: (a) K = 2π I T = 2π (0.3308)(5.97 × 1024 kg)(6.38 × 106 m) (86.164 s) = 2.14 × 1029 J (b) K = 9.78 ⎛ 2π R ⎞ 2π (5.97 × 1024 kg)(1.50 × 1011 m)2 M⎜ = = 2.66 × 1033 J ⎟ ⎝ T ⎠ (3.156 × 107 s) (c) Since the earth’s moment of inertia is less than that of a uniform sphere, more of the earth’s mass must be concentrated near its center EVALUATE: These kinetic energies are very large, because the mass of the earth is very large IDENTIFY: Using energy considerations, the system gains as kinetic energy the lost potential energy, mgR 1 SET UP: The kinetic energy is K = I ω + mv , with I = 12 mR for the disk v = Rω 2 1 4g EXECUTE: K = I ω + m(ω R ) = ( I + mR )ω Using Ι = 12 mR and solving for ω , ω = and 2 3R ω= 4g 3R EVALUATE: The small object has speed v = from a height h, it would attain a speed factor of 9.79 2 gR If it was not attached to the disk and was dropped gR Being attached to the disk reduces its final speed by a IDENTIFY: Use Eq (9.20) to calculate I Then use K = 12 I ω to calculate K (a) SET UP: The object is sketched in Figure 9.79 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9-23 Consider a small strip of width dy and a distance y below the top of the triangle The length of the strip is x = ( y/h)b Figure 9.79 EXECUTE: The strip has area x dy and the area of the sign is bh, so the mass of the strip is ⎛ x dy ⎞ ⎛ yb ⎞⎛ dy ⎞ ⎛ M ⎞ dm = M ⎜ ⎟ = M ⎜ ⎟⎜ ⎟=⎜ ⎟ y dy ⎜ bh ⎟ ⎝ h ⎠⎝ bh ⎠ ⎝ h ⎠ ⎝2 ⎠ ⎛ Mb ⎞ dI = 13 ( dm) x = ⎜ y dy ⎜ 3h ⎟⎟ ⎝ ⎠ h I = ∫ dI = Mb 3h h ∫0 y dy = Mb ⎛ ⎜ y 3h ⎝ h⎞ 0⎟= ⎠ Mb (b) I = 16 Mb = 2.304 kg ⋅ m ω = 2.00 rev/s = 4.00π rad/s K = 12 I ω = 182 J EVALUATE: From Table (9.2), if the sign were rectangular, with length b, then I = 13 Mb Our result is 9.80 one-half this, since mass is closer to the axis for the triangular than for the rectangular shape IDENTIFY: Apply conservation of energy to the system SET UP: For the falling mass K = 12 mv For the wheel K = 12 I ω EXECUTE: (a) The kinetic energy of the falling mass after 2.00 m is K = 12 mv = 12 (8.00 kg)(5.00 m/s)2 = 100 J The change in its potential energy while falling is mgh = (8.00 kg)(9.8 m/s )(2.00 m) = 156.8 J The wheel must have the “missing” 56.8 J in the form of rotational kinetic energy Since its outer rim is v 5.00 m/s = 13.51 rad/s moving at the same speed as the falling mass, 5.00 m/s , v = rω gives ω = = r 0.370 m 2K 2(56.8 J) K = I ω ; therefore I = = = 0.622 kg ⋅ m ω (13.51 rad/s) (b) The wheel’s mass is (280 N)/(9.8 m/s ) = 28.6 kg The wheel with the largest possible moment of inertia would have all this mass concentrated in its rim Its moment of inertia would be I = MR = (28.6 kg)(0.370 m) = 3.92 kg ⋅ m The boss’s wheel is physically impossible EVALUATE: If the mass falls from rest in free fall its speed after it has descended 2.00 m is v = g (2.00 m) = 6.26 m/s Its actual speed is less because some of the energy of the system is in the 9.81 form of rotational kinetic energy of the wheel IDENTIFY: Use conservation of energy The stick rotates about a fixed axis so K = 12 I ω Once we have ω use v = rω to calculate v for the end of the stick SET UP: The object is sketched in Figure 9.81 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-24 Chapter Take the origin of coordinates at the lowest point reached by the stick and take the positive y-direction to be upward Figure 9.81 EXECUTE: (a) Use Eq.(9.18): U = Mgycm ΔU = U − U1 = Mg ( ycm2 − ycm1 ) The center of mass of the meter stick is at its geometrical center, so ycm1 = 1.00 m and ycm2 = 0.50 m Then ΔU = (0.180 kg)(9.80 m/s )(0.50 m − 1.00 m) = -0.882 J (b) Use conservation of energy: K1 + U1 + Wother = K + U Gravity is the only force that does work on the meter stick, so Wother = K1 = Thus K = U1 − U = −ΔU , where ΔU was calculated in part (a) K = 12 I ω22 so Iω 2 L = 1.00 m, so ω2 = = −ΔU and ω2 = 2(−ΔU )/I For stick pivoted about one end, I = 13 ML2 where 6( −ΔU ) ML = 6(0.882 J) (0.180 kg)(1.00 m) = 5.42 rad/s (c) v = rω = (1.00 m)(5.42 rad/s) = 5.42 m/s (d) For a particle in free fall, with + y upward, v0 y = 0; y − y0 = − 1.00 m; a y = −9.80 m/s ; and v y = ? Solving the equation v 2y = v02y + 2a y ( y − y0 ) for v y gives v y = - 2a y ( y − y0 ) = - 2( −9.80 m/s )(−1.00 m) = - 4.43 m/s EVALUATE: The magnitude of the answer in part (c) is larger U1,grav is the same for the stick as for a particle falling from a height of 1.00 m For the stick K = 12 I ω22 = 12 ( 13 ML2 )(v/L) = 16 Mv For the stick and for the particle, K is the same but the same K gives a larger v for the end of the stick than for the 9.82 particle The reason is that all the other points along the stick are moving slower than the end opposite the axis IDENTIFY: Apply conservation of energy to the system of cylinder and rope SET UP: Taking the zero of gravitational potential energy to be at the axle, the initial potential energy is zero (the rope is wrapped in a circle with center on the axle).When the rope has unwound, its center of mass is a distance π R below the axle, since the length of the rope is 2π R and half this distance is the position of the center of the mass Initially, every part of the rope is moving with speed ω0 R, and when the rope has unwound, and the cylinder has angular speed ω , the speed of the rope is ω R (the upper end of the rope has the same tangential speed at the edge of the cylinder) I = (1 2) MR for a uniform cylinder ⎛M m⎞ ⎛M m⎞ EXECUTE: K1 = K + U ⎜ + ⎟ R 2ω02 = ⎜ + ⎟ R 2ω − mgπ R Solving for ω gives ⎝ ⎠ ⎝ 2⎠ ω = ω02 + (4π mg/R ) , and the speed of any part of the rope is v = ω R ( M + m) 2π g and v = v02 + 2π gR This is the R final speed when an object with initial speed v0 descends a distance π R EVALUATE: When m → 0, ω → ω0 , When m >> M , ω = ω02 + © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9.83 9-25 IDENTIFY: Apply conservation of energy to the system consisting of blocks A and B and the pulley SET UP: The system at points and of its motion is sketched in Figure 9.83 Figure 9.83 Use the work-energy relation K1 + U1 + Wother = K + U Use coordinates where + y is upward and where the origin is at the position of block B after it has descended The tension in the rope does positive work on block A and negative work of the same magnitude on block B, so the net work done by the tension in the rope is zero Both blocks have the same speed EXECUTE: Gravity does work on block B and kinetic friction does work on block A Therefore Wother = W f = − μ k m A gd K1 = (system is released from rest) U1 = mB gy B1 = mB gd ; U = mB gyB = K = 12 m Av22 + 12 mB v22 + 12 I ω22 But v(blocks) = Rω (pulley), so ω2 = v2 /R and K = 12 ( mA + mB )v22 + 12 I (v2 /R )2 = 12 (m A + mB + I/R )v22 Putting all this into the work-energy relation gives mB gd − μ k m A gd = 12 (m A + mB + I/R )v22 (m A + mB + I/R )v22 = gd (mB − μ k m A ) v2 = gd (mB − μ k m A ) m A + mB + I/R EVALUATE: If mB >> m A and I/R , then v2 = gd ; block B falls freely If I is very large, v2 is very small Must have mB > μk m A for motion, so the weight of B will be larger than the friction force on A 9.84 I/R has units of mass and is in a sense the “effective mass” of the pulley IDENTIFY: Apply conservation of energy to the system of two blocks and the pulley SET UP: Let the potential energy of each block be zero at its initial position The kinetic energy of the system is the sum of the kinetic energies of each object v = Rω , where v is the common speed of the blocks and ω is the angular velocity of the pulley EXECUTE: The amount of gravitational potential energy which has become kinetic energy is K = (4.00 kg − 2.00 kg)(9.80 m/s )(5.00 m) = 98.0 J In terms of the common speed v of the blocks, the 1 ⎛v⎞ kinetic energy of the system is K = (m1 + m2 )v + I ⎜ ⎟ 2 ⎝R⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-26 Chapter 1⎛ (0.560 kg ⋅ m ) ⎞ K = v ⎜ 4.00 kg + 2.00 kg + ⎟ = v (13.94 kg) Solving for v gives ⎜⎝ (0.160 m) ⎟⎠ v= 98.0 J = 2.65 m/s 13.94 kg EVALUATE: If the pulley is massless, 98.0 J = 12 (4.00 kg + 2.00 kg)v and v = 5.72 m/s The moment of 9.85 inertia of the pulley reduces the final speed of the blocks IDENTIFY and SET UP: Apply conservation of energy to the motion of the hoop Use Eq (9.18) to calculate U grav Use K = 12 I ω for the kinetic energy of the hoop Solve for ω The center of mass of the hoop is at its geometrical center Take the origin to be at the original location of the center of the hoop, before it is rotated to one side, as shown in Figure 9.85 Figure 9.85 ycm1 = R − R cos β = R(1 − cos β ) ycm2 = (at equilibrium position hoop is at original position) EXECUTE: K1 + U1 + Wother = K + U Wother = (only gravity does work) K1 = (released from rest), K = 12 I ω22 For a hoop, I cm = MR , so I = Md + MR with d = R and I = 2MR , for an axis at the edge Thus K = 12 (2 MR )ω22 = MR 2ω22 U1 = Mgycm1 = MgR(1 − cos β ), U = mgycm2 = Thus K1 + U1 + Wother = K + U gives MgR (1 − cos β ) = MR 2ω22 and ω2 = g (1 − cos β )/R If β = 0, then ω2 = As β increases, ω2 increases EVALUATE: 9.86 IDENTIFY: K = 12 I ω , with ω in rad/s P = energy t SET UP: For a solid cylinder, I = 12 MR rev/min = (2π /60) rad/s EXECUTE: (a) ω = 3000 rev/min = 314 rad/s I = 12 (1000 kg)(0.900 m) = 405 kg ⋅ m K = 12 (405 kg ⋅ m )(314 rad/s)2 = 2.00 × 107 J (b) t = K 2.00 × 107 J = = 1.08 × 103 s = 17.9 P 1.86 × 104 W EVALUATE: In K = 12 I ω , we must use ω in rad/s 9.87 IDENTIFY: I = I1 + I Apply conservation of energy to the system The calculation is similar to Example 9.8 SET UP: ω = v v for part (b) and ω = for part (c) R1 R2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9-27 1 EXECUTE: (a) I = M1R12 + M R22 = ((0.80 kg)(2.50 × 10−2 m) + (1.60 kg)(5.00 × 10−2 m) ) 2 I = 2.25 × 10−3 kg ⋅ m (b) The method of Example 9.8 yields v = v= gh + ( I/mR12 ) 2(9.80 m/s )(2.00 m) (1 + ((2.25 × 10−3 kg ⋅ m )/(1.50 kg)(0.025 m) )) = 3.40 m/s (c) The same calculation, with R2 instead of R1 gives v = 4.95 m/s EVALUATE: The final speed of the block is greater when the string is wrapped around the larger disk v = Rω , so when R = R2 the factor that relates v to ω is larger For R = R2 a larger fraction of the total kinetic energy resides with the block The total kinetic energy is the same in both cases (equal to mgh), so when R = R2 the kinetic energy and speed of the block are greater 9.88 IDENTIFY: The potential energy of the falling block is transformed into kinetic energy of the block and kinetic energy of the turning wheel, but some of it is lost to the work by friction Energy conservation applies, with the target variable being the angular velocity of the wheel when the block has fallen a given distance SET UP: K1 + U1 + Wother = K + U , where K = mv , U = mgh, and Wother is the work done by friction 1 1 EXECUTE: Energy conservation gives mgh + (−6.00 J) = mv + I ω v = Rω , so mv = mR 2ω 2 2 and mgh + (−6.00 J) = (mR + I )ω Solving for ω gives ω= 9.89 2[mgh + (−6.00 J)] mR + I = 2[(0.340 kg)(9.8 m/s )(3.00 m) − 6.00 J] (0.340 kg)(0.180 m) + 0.480 kg ⋅ m = 4.03 rad/s EVALUATE: Friction does negative work because it opposes the turning of the wheel IDENTIFY: Apply conservation of energy to relate the height of the mass to the kinetic energy of the cylinder SET UP: First use K (cylinder) = 480 J to find ω for the cylinder and v for the mass EXECUTE: I = 12 MR = 12 (10.0 kg)(0.150 m) = 0.1125 kg ⋅ m K = 12 I ω so ω = K/I = 92.38 rad/s v = Rω = 13.86 m/s SET UP: Use conservation of energy K1 + U1 = K + U to solve for the distance the mass descends Take y = at lowest point of the mass, so y2 = and y1 = h, the distance the mass descends EXECUTE: K1 = U = so U1 = K mgh = 12 mv + 12 I ω , where m = 12.0 kg For the cylinder, I = 12 MR and ω = v/R, so h= Iω 2 = 14 Mv Solving mgh = 12 mv + 14 Mv for h gives v2 ⎛ M ⎞ ⎜1 + ⎟ = 13.9 m g ⎝ 2m ⎠ EVALUATE: For the cylinder K cyl = 12 I ω = 12 ( 12 MR )(v/R) = 14 Mv K mass = 12 mv , so K mass = (2m /M ) K cyl = [2(12.0 kg)/10.0 kg](480 J) = 1150 J The mass has 1150 J of kinetic energy when the cylinder has 480 J of kinetic energy and at this point the system has total energy 1630 J since U = Initially the total energy of the system is U1 = mgy1 = mgh = 1630 J, so the total energy is shown to be conserved © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-28 9.90 Chapter IDENTIFY: Energy conservation: Loss of U of box equals gain in K of system Both the cylinder and pulley have kinetic energy of the form K = 12 I ω 1 2 mbox gh = mbox vbox + I pulleyωpulley + I cylinderωcylinder 2 v v SET UP: ωpulley = box and ωcylinder = box rp rcylinder Let B = box, P = pulley and C = cylinder 2 1⎛1 ⎞⎛v ⎞ ⎛1 ⎞⎛v ⎞ EXECUTE: mB gh = mBvB2 + ⎜ mP rP2 ⎟ ⎜ B ⎟ + ⎜ mC rC2 ⎟ ⎜ B ⎟ 2⎝2 ⎠ ⎝ rP ⎠ ⎝ ⎠ ⎝ rC ⎠ mB gh = vB = 1 mBvB2 + mP vB2 + mCvB2 and 4 mB gh (3.00 kg)(9.80 m/s )(2.50 m) = = 4.76 m/s 1m + 1m + 1m 1.50 kg + 14 (7.00 kg) B P C EVALUATE: If the box was disconnected from the rope and dropped from rest, after falling 2.50 m its speed would be v = g (2.50 m) = 7.00 m/s Since in the problem some of the energy of the system goes 9.91 into kinetic energy of the cylinder and of the pulley, the final speed of the box is less than this IDENTIFY: I = I disk − I hole , where I hole is I for the piece punched from the disk Apply the parallel-axis theorem to calculate the required moments of inertia SET UP: For a uniform disk, I = 12 MR EXECUTE: (a) The initial moment of inertia is I = 12 MR The piece punched has a mass of M and a 16 moment of inertia with respect to the axis of the original disk of 2 M ⎡1 ⎛ R ⎞ ⎛ R ⎞ ⎤ MR ⎢ ⎜ ⎟ +⎜ ⎟ ⎥ = 16 ⎢⎣ ⎝ ⎠ ⎝ ⎠ ⎥⎦ 512 247 The moment of inertia of the remaining piece is then I = MR − MR = MR 512 512 383 (b) I = 12 MR + M ( R / 2) − 12 ( M /16)( R /4) = 512 MR EVALUATE: For a solid disk and an axis at a distance R /2 from the disk’s center, the parallel-axis theorem gives I = 12 MR = 34 MR = 384 MR For both choices of axes the presence of the hole reduces I, 512 9.92 but the effect of the hole is greater in part (a), when it is farther from the axis IDENTIFY: We know (or can calculate) the masses and geometric measurements of the various parts of the body We can model them as familiar objects, such as uniform spheres, rods, and cylinders, and calculate their moments of inertia and kinetic energies SET UP: My total mass is m = 90 kg I model my head as a uniform sphere of radius cm I model my trunk and legs as a uniform solid cylinder of radius 12 cm I model my arms as slender rods of length 60 cm ω = 72 rev/min = 7.5 rad/s For a solid uniform sphere, I = 2/5 MR2, for a solid cylinder, I = 12 MR , and for a rod rotated about one end I = 1/3 ML2 EXECUTE: (a) Using the formulas indicated above, we have Itot = Ihead + Itrunk+legs + Iarms, which gives I tot = 52 (0.070m)(0.080 m) + 12 (0.80m)(0.12 m) + 13 (0.13m)(0.60 m) = 3.3 kg ⋅ m where we have () used m = 90 kg (b) K rot = 12 I ω = 12 (3.3 kg ⋅ m )(7.5 rad/s)2 = 93 J EVALUATE: According to these estimates about 85% of the total I is due to the outstretched arms If the initial translational kinetic energy 12 mv of the skater is converted to this rotational kinetic energy as he goes into a spin, his initial speed must be 1.4 m/s © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies 9.93 9-29 IDENTIFY: The total kinetic energy of a walker is the sum of his translational kinetic energy plus the rotational kinetic of his arms and legs We can model these parts of the body as uniform bars SET UP: For a uniform bar pivoted about one end, I = 13 mL2 v = 5.0 km/h = 1.4 m/s K tran = 12 mv and K rot = 12 I ω EXECUTE: (a) 60° = ( 13 ) rad The average angular speed of each arm and leg is rad 1s = 1.05 rad/s (b) Adding the moments of inertia gives I = 13 marm Larm + 13 mleg Lleg = 13 [(0.13)(75 kg)(0.70 m)2 + (0.37)(75 kg)(0.90 m) ] I = 9.08 kg ⋅ m K rot = 12 I ω = 12 (9.08 kg ⋅ m )(1.05 rad/s)2 = 5.0 J (c) K tran = 12 mv = 12 (75 kg)(1.4 m/s) = 73.5 J and K tot = K tran + K rot = 78.5 J (d) 9.94 K rot 5.0 J = = 6.4% K tran 78.5 J EVALUATE: If you swing your arms more vigorously more of your energy input goes into the kinetic energy of walking and it is more effective exercise Carrying weights in our hands would also be effective IDENTIFY: The total kinetic energy of a runner is the sum of his translational kinetic energy plus the rotational kinetic of his arms and legs We can model these parts of the body as uniform bars SET UP: Now v = 12 km/h = 3.33 m/s I tot = 9.08 kg ⋅ m as in Problem 9.93 EXECUTE: (a) ωav = 1/3 rad = 2.1 rad/s 0.5 s (b) K rot = 12 I ω = 12 (9.08 kg ⋅ m )(2.1 rad/s) = 20 J (c) K tran = 12 mv = 12 (75 kg)(3.33 m/s) = 416 J Therefore K tot = K tran + K rot = 416 J + 20 J = 436 J (d) 9.95 K rot 20 J = = 4.6% K tot 436 J IDENTIFY: Follow the instructions in the problem to derive the perpendicular-axis theorem Then apply that result in part (b) SET UP: I = ∑ mi ri2 The moment of inertia for the washer and an axis perpendicular to the plane of the i washer at its center is center is M 12 EXECUTE: IO = M (R2 (L + L ) = 2 + R22 ) In part (b), I for an axis perpendicular to the plane of the square at its ML2 (a) With respect to O, ri2 = xi2 + yi2 , and so ∑ mi ri2 =∑ mi ( xi2 + yi2 ) = ∑ mi xi2 +∑ mi yi2 = I x + I y i i i i (b) Two perpendicular axes, both perpendicular to the washer’s axis, will have the same moment of inertia about those axes, and the perpendicular-axis theorem predicts that they will sum to the moment of inertia about the washer axis, which is I = 12 M ( R12 + R22 ), and so I x = I y = 14 M ( R12 + R22 ) (c) I = 16 mL2 Since I = I x + I y , and I x = I y , both I x and I y must be 9.96 12 mL2 EVALUATE: The result in part (c) says that I is the same for an axis that bisects opposite sides of the square as for an axis along the diagonal of the square, even though the distribution of mass relative to the two axes is quite different in these two cases IDENTIFY: Apply the parallel-axis theorem to each side of the square SET UP: Each side has length a and mass M/4, and the moment of inertia of each side about an axis perpendicular to the side and through its center is ( 1 Ma 12 )= Ma 48 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-30 Chapter EXECUTE: The moment of inertia of each side about the axis through the center of the square is, from the perpendicular axis theorem, Ma M ⎛ a ⎞ Ma + ⎜ ⎟ = The total moment of inertia is the sum of the 48 ⎝2⎠ 12 Ma Ma = 12 EVALUATE: If all the mass of a side were at its center, a distance a/2 from the axis, we would have contributions from the four sides, or × ⎛ M ⎞⎛ a ⎞ I = ⎜ ⎟⎜ ⎟ = Ma If all the mass was divided equally among the four corners of the square, a 4 ⎝ ⎠⎝ ⎠ 9.97 ⎛ M ⎞⎛ a ⎞ distance a/ from the axis, we would have I = ⎜ ⎟⎜ ⎟ = Ma The actual I is between these ⎝ ⎠⎝ ⎠ two values IDENTIFY: Use Eq (9.20) to calculate I (a) SET UP: Let L be the length of the cylinder Divide the cylinder into thin cylindrical shells of inner radius r and outer radius r + dr An end view is shown in Figure 9.97 ρ = αr The mass of the thin cylindrical shell is dm = ρ dV = ρ (2π r dr ) L = 2πα Lr dr Figure 9.97 ( R ) = πα LR R M = ∫ dm = 2πα L ∫ r dr = 2απ L ( R ) = πα LR , so πα LR R EXECUTE: I = ∫ r dm = 2πα L ∫ r dr = 2πα L Relate M to α : 5 5 3 3 = 3M/2 Using this in the above result for I gives I = 52 (3M/2) R = 53 MR (b) EVALUATE: For a cylinder of uniform density I = 12 MR The answer in (a) is larger than this Since 9.98 the density increases with distance from the axis the cylinder in (a) has more mass farther from the axis than for a cylinder of uniform density IDENTIFY: Write K in terms of the period T and take derivatives of both sides of this equation to relate dK/dt to dT/dt 2π SET UP: ω = and K = 12 I ω The speed of light is c = 3.00 × 108 m/s T dK 4π I dT 4π I dT The rate of energy loss is Solving for the =- dt T T dt T dt moment of inertia I in terms of the power P, EXECUTE: (a) K = I= (b) R = (c) v = 2π I PT 4π 2 (5 × 1031 W)(0.0331 s)3 1s = = 1.09 × 1038 kg ⋅ m −13 dT/dt 4π 4.22 × 10 s 5I 5(1.08 × 1038 kg ⋅ m ) = = 9.9 × 103 m, about 10 km 2M 2(1.4)(1.99 × 1030 kg) 2π R 2π (9.9 × 103 m) = = 1.9 × 106 m/s = 6.3 × 10−3 c T (0.0331 s) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Rotation of Rigid Bodies (d) ρ = 9.99 9-31 M M = = 6.9 × 1017 kg/m3 , which is much higher than the density of ordinary rock by 14 V ( 4π /3) R3 orders of magnitude, and is comparable to nuclear mass densities EVALUATE: I is huge because M is huge A small rate of change in the period corresponds to a large release of energy IDENTIFY: The density depends on the distance from the center of the sphere, so it is a function of r We need to integrate to find the mass and the moment of inertia SET UP: M = ∫ dm = ∫ ρ dV and I = ∫ dI EXECUTE: (a) Divide the sphere into thin spherical shells of radius r and thickness dr The volume of each shell is dV = 4π r dr ρ (r ) = a − br , with a = 3.00 × 103 kg/m3 and b = 9.00 × 103 kg/m Integrating R ⎞ ⎛ gives M = ∫ dm = ∫ ρ dV = ∫ (a − br )4π r dr = π R3 ⎜ a − bR ⎟ ⎠ ⎝ ⎛ ⎞ M = π (0.200)3 ⎜ 3.00 × 103 kg/m3 − (9.00 × 103 kg/m )(0.200 m) ⎟ = 55.3 kg ⎝ ⎠ (b) The moment of inertia of each thin spherical shell is 2 8π dI = r 2dm = r ρ dV = r (a − br )4π r 2dr = r ( a − br )dr 3 3 R 8π R 8π ⎛ 5b ⎞ I = ∫ dI = r (a − br ) dr = R ⎜ a − R ⎟ ∫0 15 ⎝ ⎠ I= 8π ⎛ ⎞ (0.200 m)5 ⎜ 3.00 × 103 kg/m3 − (9.00 × 103 kg/m )(0.200 m) ⎟ = 0.804 kg ⋅ m 15 ⎝ ⎠ EVALUATE: We cannot use the formulas M = ρV and I = MR because this sphere is not uniform throughout Its density increases toward the surface For a uniform sphere with density 3.00 × 103 kg/m3 , the mass is π R3ρ = 100.5 kg The mass of the sphere in this problem is less than this For a uniform sphere with mass 55.3 kg and R = 0.200 m, I = MR = 0.885 kg ⋅ m The moment of inertia for the sphere in this problem is less than this, since the density decreases with distance from the center of the sphere 9.100 IDENTIFY: Apply Eq (9.20) SET UP: Let z be the coordinate along the vertical axis r ( z ) = EXECUTE: I = ∫ dI = πρ R ∫ h h4 z dz = R2 z2 πρ R 4 zR dm = πρ and dI = z dz h h4 h πρ R ⎡ ⎤ h = πρ R 4h The volume of a right circular cone is z ⎣ ⎦ 10 h V = 13 π R h, the mass is 13 πρ R h and so I = 10 ⎛ πρ R h ⎞ ⎜ ⎟ R = MR 10 ⎜⎝ ⎟⎠ 10 EVALUATE: For a uniform cylinder of radius R and for an axis through its center, I = 12 MR I for the 9.101 cone is less, as expected, since the cone is constructed from a series of parallel discs whose radii decrease from R to zero along the vertical axis of the cone IDENTIFY: Follow the steps outlined in the problem SET UP: ω z = dθ /dt α z = d 2ω z /dt β EXECUTE: (a) ds = r dθ = r0dθ + βθ dθ so s (θ ) = r0θ + θ θ must be in radians β (b) Setting s = vt = r0θ + θ gives a quadratic in θ The positive solution is © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 9-32 Chapter θ (t ) = 1⎡ r0 + 2β vt − r0 ⎤ ⎥⎦ β ⎢⎣ (The negative solution would be going backwards, to values of r smaller than r0 ) (c) Differentiating, ω z (t ) = dθ v dω z β v2 = , αz = =- The angular acceleration α z dt dt (r0 + 2β vt )3 / r02 + 2β vt is not constant (d) r0 = 25.0 mm θ must be measured in radians, so β = (1.55μ m/rev)(1 rev/2π rad) = 0.247 μ m/rad Using θ (t ) from part (b), the total angle turned in 74.0 = 4440 s is θ= 2.47 × 10 -7 ( m/rad 2(2.47 × 10-7 m/rad)(1.25 m/s)(4440 s) + (25.0 × 10-3 m)2 − 25.0 × 10-3 m ) θ = 1.337 × 105 rad, which is 2.13 × 104 rev (e) The graphs are sketched in Figure 9.101 EVALUATE: ω z must decrease as r increases, to keep v = rω constant For ω z to decrease in time, α z must be negative Figure 9.101 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... arm and leg is rad 1s = 1.05 rad/s (b) Adding the moments of inertia gives I = 13 marm Larm + 13 mleg Lleg = 13 [(0 .13) (75 kg)(0.70 m)2 + (0.37)(75 kg)(0.90 m) ] I = 9.08 kg ⋅ m K rot = 12 I ω... − β t = implies t = 2γ γ 3.20 rad/s = = = 2 .133 s β 3β 3(0.500 rad/s3 ) At this t, ω z = 2γ t − 3β t = 2(3.20 rad/s )(2 .133 s) − 3(0.500 rad/s3 )(2 .133 s) = 6.83 rad/s 9.65 The maximum positive... = − 0 .139 rad/s3 (b) θ = π /4 rad and α z = at t = (c) α z = 3.50 rad/s at t = − θ= π αz 6c =− 3.50 rad/s 6( − 0 .139 rad/s3 ) = 4.20 s At t = 4.20 s, rad + (2.00 rad/s)(4.20 s) − (− 0 .139 rad/s3