Solution manual for vibrations 2nd edition by balachandran

Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Solutions to Exercises Chapter Section 1.1 1.1 Choose any two contributors from Table 1.1, study their contributions, and write a paragraph about each of them Section 1.2.1 1.2 Consider the planar pendulum kinematics discussed in Example 1.1, start with position vector r P O resolved in terms the unit vectors i and j, and verify the expressions obtained for the acceleration and velocity given by Eq (f) of Example 1.1 Solution 1.2 The relationships among the unit vectors are e1  cos i  sin  j e2   sin  i  cos j Then, the position vector is r P / O  rQ /O  r P /Q  hj  Le2  L sin  i  (h  L cos ) j The velocity vector is dr P O d d  ( L sin  i )   ( h  L cos  ) j  dt dt dt  L cos i  L sin  j  L  cos i  L sin  j  vP /O   L e1 The acceleration vector is aP /O  dv P / O d d  ( L cos  i )  ( L sin  j ) dt dt dt       L sin   L cos i  L cos  L sin  j  L (  sin  i  cos j )  L (cos i  sin  j )  L e1  L e2 1.3 Consider the kinematics of the rolling disc considered in Example 1.2, and verify that the instantaneous acceleration of the point of contact is not zero ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Solution 1.3 Making use of Eq (1.7) that relates accelerations of two arbitrary points on a rigid body, we have aC / O  aG / O      r C / G     r C / G From Example 1.2, we have r C / G   rj aG / O  r i    k    k and, hence,      aC / O  r i   k   k    rj    k    rj   r i  r j  r i  r j Thus, it is verified that the instantaneous acceleration of the point of contact is not zero 1.4 Show that the acceleration of the particle in the rotating frame of Example 1.3 is a  ( x p  2 y p   x p   y p )e1  ( y p  2 x p   y p   x p )e2 where  is the magnitude of the angular acceleration of the rotating frame about the z axis Solution 1.4 From Eq (b) of Example 1.3 v dr  ( x p   y p )e1  ( y p   x p )e2 dt Then the acceleration is dv d d  ( x p   y p )e1   ( y p   x p )e2  dt dt dt d de d de  e1 ( x p   y p )  ( x p   y p )  e2 ( y p   x p )  ( y p   x p ) dt dt dt dt  ( x p   y p   y p )e1  ( x p   y p )(  e1 )  ( y p   x p   x p )e2  ( y p   x p )(  e2 ) a where  = d/dt and we have used Eq (1.8) We note that  = k; therefore, ω  e1   k  e1   e2 ω  e2   k  e2   e1 Then ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter a  ( x p  2 y p   x p   y p )e1  ( y p  2 x p   y p   x p )e2 1.5 In Figure E1.5, a slider of mass Mr is located on a bar whose angular displacement in the plane is described by the coordinate  The motion of the slider from the pivot point is measured by the coordinate r1 The acceleration due to gravity acts in a direction normal to the plane of motion Assume that the point O is fixed in an inertial reference frame and determine the absolute velocity and absolute acceleration of the slider Solution 1.5 Choose unit vectors e1 and e2 fixed to the slider as shown in the figure Then, the position vector from point O to the slider is e2 e1 r1 rm  r1e1 The absolute velocity is then given by Vm  d de  r1e1   r1e1  r1  r1e1  r1   e1  dt dt (t) O Noting that    k , where k is the unit vector that points out of the plane, we find that Vm  r1e1  r1 k  e1  r1e1  r1 e2 The absolute acceleration is found from d de  de  r1e1  r1 e2  r1e1  r1  r1 e2  r1 e2  r1 dt dt dt  r1e1  r1   e1   r1 e2  r1 e2  r1   e2   am        r1  r1 e1  r1  2r1 e2 1.6 A pendulum of mass m is attached to a moving pivot of mass M as shown in Figure E1.6 Assume that the pivot point cannot translate in the vertical direction If the horizontal translation of the pivot point from the fixed point O is measured by the coordinate x and the angle  is used to describe the angular displacement of the pendulum from the vertical, determine the absolute velocity of the pendulum ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Solution 1.6 The position vector of the location of the mass with respect to the fixed-point O is given by j M i rm  r O O  r m O O O x rm  xi  le2 Then the velocity is dr Vm  m  xi  le2  xi  l  ω  e2   xi  l (  k )  e2 dt  xi  l e1  xi  l  cos  i  sin  j   l eˆ2 eˆ1 m   x  l cos   i  l sin  j We could have also obtained the solution as follows: rm  xi  l sin  i  l cos  j   x  l sin   i  l cos  j Then the velocity is Vm  rm   x  l cos  i  l sin  j Section 1.2.2 1.7 Determine the number of degrees of freedom for the systems shown in Figure E1.7 Assume that the length L of the pendulum shown in Figure E1.7a is constant and that the length between each pair of particles in Figure E1.7b is constant Hint: For Figure E1.7c, the rigid body can be thought of as a system of particles where the length between each pair of particles is constant Solution 1.7 a) Three coordinates (n = 3) and one inextensible constraint (m = 1): number of degrees of freedom is  = b) Three particles, each with coordinates and three inextensible constraints: number of degrees of freedom is  = c) Rigid body: number of degrees of freedom is 6; translation and rotational The system of three particles shown in Figure E1.7b is an example of a rigid body, since the particles are a constant distance apart from each other In three-dimensional space, for a system of N particles representing a rigid body, the number of degrees of freedom is 3N  (3N  6), where 3N is the total number of coordinates associated with the N particles and 3N  is the total number of inextensional constraints that ensures that the particles are always at a constant distance apart By extension to the case shown in Figure E1.7c, the number of degrees of freedom is ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Section 1.2.3 1.8 Draw free-body diagrams for each of the masses shown in Figure E1.6 and obtain the equations of motion along the horizontal direction by using Eq (1.15) Solution 1.8 The free-body diagrams for each of the masses in figure E1.14 are as shown to the right The force N is shown in the free-body diagram of mass M to account for the constraint that the pivot mass cannot move in the j-direction Summing forces along the i-direction for each of the two masses, we find for the pivot mass from Eqs (1.15) that N j T sin   Mx M i T  Mg m and for the pendulum mass that T sin   m  x  l sin   l cos  mg 1.9 Draw the free-body diagram for the whole system shown in Figure E1.6, obtain the system equation of motion by using Eq (1.14) along the horizontal direction, and verify that this equation can be obtained from Eq (1.15) Solution 1.9 The free-body diagram is shown to the right Note that the internal force T does not appear in this diagram Making use of Eqs (1.14) for the motion along the horizontal direction, we find that  Mx  m  x  l sin   l cos  which can be obtained by adding the two equations obtained to the solution to Exercise 1.8 N M  Mg m mg 1.10 Determine the linear momentum for the system shown in Figure E1.5 and discuss if it is conserved Assume that the mass of the bar is Mbar and the distance from the point O to the center of the bar is Lbar Solution 1.10 The linear momentum of the system is given by p  pbar  pslider where pbar  M bar Lbar e2 ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Making use of the slider velocity Vm determined in the solution to Exercise 1.5, we arrive at  pslider  M r r1e1  r1 e2  Thus, p  M r r1e1   M bar Lbar  M r r1  e2 Since there are no external forces acting on the system, by virtue of Eq (1.14), the system’s linear momentum is conserved 1.11 Determine the angular momentum of the system shown in Figure 1.6 about the point O and discuss if it is conserved Solution 1.11 The angular momentum of the system about the point O is given by H  H pendulum  H disc  J O k  rm  p where k is the unit vector normal to the plane, the position vector rm runs from point O to the pendulum, and p is the linear momentum of the pendulum Making use of the solution to Example 1.4, we have rm   R  r cos   e1  r sin  e2         p  mVm  m  r    sin  e1  R  r    cos  e2 and       H   J O   R  r cos  R  r    cos  r sin      k   which can be written as     H   J O  R 2  r     rR 2   cos  k   If no external moments act on the system, since O is a fixed point, from Eq (1.17) it is clear that the angular momentum of the system is conserved 1.12 A rigid body is suspended from the ceiling by two elastic cables that are attached to the body at the points O and O, as shown in Figure E1.12 Point G is the center of mass of the body Which of these points would you choose to carry out an angular-momentum balance based on Eq (1.17)? Solution 1.12 Since point G is the center of mass of the body, this point would be used to carry out an angular momentum balance based on Eq (1.17) One cannot use Eq (1.17) with points O and O since neither of them is a fixed point ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter 1.13 Consider the rigid body shown in Figure E1.13 This body has a mass m and rotary inertia JG about the center of mass G It is suspended from a point O on the ceiling by using an elastic suspension The point of attachment O is at a distance l from the center of mass G of this body M(t) is an external moment applied to the system along an axis normal to the plane of the body Use the generalized coordinates x, which describes the up and down motions of point O from point O, and , which describes the angular oscillations about an axis normal to the plane of the rigid body For the system shown in Figure E1.13, use the principle of angular-momentum balance given by Eq (1.17) and obtain an equation of motion for the system Assume that gravity loading is present Solution 1.13 The principle of angular-momentum balance given by Eq (1.17) is applied with respect to the center of mass of the system Thus,  M (t ) k   d J G k dt  where k is the unit vector normal to the plane of the system Then we have  M (t )  J G If, instead, the principle of angular-momentum balance given by Eq (1.17) is applied with respect to the fixed point O of the system, the result is  M (t ) k   x  l sin   i  l cos  j    mgi    d J G k   x  l sin   i  l cos  j    mVm  dt d  J G k   x  l sin   i  l cos  j    m x  l cos  i  l sin  j    dt      where, we have substituted the velocity of the center of mass from the solution to Exercise 1.13 Carrying out the different cross-product operations, we obtain   M (t )  mgl cos  k  J G k  m     d l  x  l sin   sin   l x  l cos cos k dt which leads to  M (t )  mgl cos  J G  ml d dt  x  l sin  sin   x  l cos  cos  Section 1.2.4 1.14 For the system shown in Figure E1.13, construct the system kinetic energy ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Solution 1.14 We make use of Eq (1.23) to determine the system kinetic energy Note that the velocity of the center of mass G can be obtained as in the solution to Exercise 1.6; that is,   Vm  x  l cos i  l sin  j Then, from Eq (1.23), the kinetic energy is 1 T  m Vm  Vm   J G 2 1  m x  l 2  xl cos  J G 2  O j m, JG i G O l   1.15 Determine the kinetic energy of the planar pendulum of Example 1.1 Solution 1.15 From Eqs (b) and (e) of Example 1.1, we find that v  Lθe1  Lθ  cos i  sin  j  Then, from Eq (1.22), the kinetic energy is T 1 m(v  v )  mL2θ  cos2   sin    mL2θ 2 2 1.16 Consider the disc rolling along a line in Figure E1.16 The disc has a mass m and a rotary inertia JG about the center of mass G Answer the following: (a) How many degrees of freedom does this system have? and (b) Determine the kinetic energy for this system Solution 1.16 a) One degree of freedom, since one independent coordinate (x or ) is needed to describe the motion Due to the non-slip constraint, x = r, and x or  can always be expressed in terms of the other coordinate b) Making use of Eq (1.24), we have T 1 m (vG  vG )  J G 2 Noting from Example 1.2 that vG  vG O  r i we arrive at T 1 m ( r i  r i )  J G   mr  J G  2 2 ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter 1.17 In the system shown in Figure 1.6, if the mass of the pendulum is m, the length of the pendulum is r, and the rotary inertia of the disc about the point O is JO, determine the system kinetic energy Solution 1.17 The kinetic energy of the system can be written as T  Tpendulum  Tdisc where m (Vm  Vm )  J O 2 Tpendulum  Tdisc Noting from the solution to Example 1.4 that       Vm  r    sin  e1  R  r    cos e2 we determine that      1 J O  m r    2 1  J O  m r    2 Tpendulum      sin   R  r    cos   R 2  2rR    cos      1.18 Referring to Figure E1.6 and assuming that the bar to which the pendulum mass m is connected is massless, determine the kinetic energy for the system Solution 1.18 The kinetic energy of the system is given by T  TM  Tpendulum where m Vm  Vm  1 TM  M  xi  xi   Mx 2 Tpendulum  Making use of the solution for the velocity in Exercise 1.6, we find that m   x  l cos   i  l sin  j     x  l cos   i  l sin  j  2  m  x  l cos    l 2 sin     Tpendulum  ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra Solutions to Exercises – Chapter Thus,  m  x  l cos    l 2 sin    Mx  2  1   m  M  x  m  l 2  xl cos   2 T 1.19 Determine the kinetic energy of the system shown in Figure E1.5 Solution 1.19 The kinetic energy of the system is given by T  Tbar  Tslider where M r (Vm  Vm )  J O 2 Tslider  Tbar Making use of the velocity determined in the solution of Exercise 1.5, we find that Tslider      1 M r r1e1  r1 e2  r1e1  r1 e2  M r r12  r12 2  and, therefore, T   1 1 J O  M r r12  r12   J O  M r r12   M r r12 2 2 10 ©2009 Cengage Learning ALL RIGHTS RESERVED Full file at https://TestbankDirect.eu/Solution-Manual-for-Vibrations-2nd-Edition-by-Balachandra ... https://TestbankDirect.eu /Solution- Manual- for- Vibrations- 2nd- Edition- by- Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu /Solution- Manual- for- Vibrations- 2nd- Edition- by- Balachandra... https://TestbankDirect.eu /Solution- Manual- for- Vibrations- 2nd- Edition- by- Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu /Solution- Manual- for- Vibrations- 2nd- Edition- by- Balachandra... https://TestbankDirect.eu /Solution- Manual- for- Vibrations- 2nd- Edition- by- Balachandra Solution Manual for Vibrations 2nd Edition by Balachandran Full file at https://TestbankDirect.eu /Solution- Manual- for- Vibrations- 2nd- Edition- by- Balachandra