Chapter 10 Solutions

13,23,30,45,65,75; 18,24,31,46,92,105; 79


13.   The final angular speed is
      w = (1 rpm)(2p rad/rev)/(60 s/min) = 0.105 rad/s.
   (a)   We find the angular acceleration from
         a    = Dw/Dt 
            = (0.105 rad/s - 0)/(10.0 min)(60 s/min) =       1.75 x10-4rad/s2.
   (b)   We find the angular speed after 5.0 min:
         w = w0 + at = 0 + (1.75 x10-4rad/s2)(5.0 min)(60 s/min) = 5.25 x10-2rad/s.
      At this time the radial acceleration of a point on the skin is
         aR = w2r = (5.25 x10-2rad/s)2(4.25 m) =       1.17 x10-2m/s2.
      The tangential acceleration is
         atan = ar = (1.75 x10-4rad/s2)(4.25 m) =       7.44 x10-4m/s2.


23.   We assume clockwise motion, so the frictional torque is counterclockwise.  
   If we take the clockwise direction as positive, we have
      tnet    = rF1 - RF2 + RF3 - tfr 
         = (0.10 m)(35 N) - (0.20 m)(30 N) + (0.20 m)(20 N) - 0.30 m  N 
         =       1.2 m  N (clockwise).


30.   (a)   For the moment of inertia about the y-axis, we have
         Ia    =  SmiRi2 = md12 + Md12 + m(d2 - d1)2 + M(d2 - d1)2 
            = (1.8 kg)(0.50 m)2  + (3.1 kg)(0.50 m)2 + 
               (1.8 kg)(1.00 m)2 + (3.1 kg)(1.00 m)2 =      6.1 kg  m2. 
   (b)   For the moment of inertia about the x-axis, all the masses 
      are the same distance from the axis, so we have
         Ib    =  SmiRi2 = (2m + 2M)(1/2h)2  
            = [2(1.8 kg) + 2(3.1 kg)](0.25 m)2  =      0.61 kg  m2. 
   It will be harder to accelerate the array around the y-axis, because the moment of inertia is greater.


45.   (a)   We use the parallel-axis theorem:
         I = 2(ICM + Mh2) = 2[(2MR02/5) + M(3R0/2)2] =       5.30MR02.
   (b)   If we treat the spheres as point masses, we get
         I' = 2[M(3R0/2)2] = 4.50MR02.
      The error is 

         error = (I' - I)/I = (4.50 - 5.30)/(5.30) = - 0.15 =      - 15%.



65.   For the system of the two blocks and pulley, no work will be 
   done by nonconservative forces.  The rope ensures that each 
   block has the same speed v and the angular speed of the 
   pulley is w = v/R0.  We choose the reference level for 
   gravitational potential energy at the floor.  
   The rotational inertia of the pulley is I = 1/2MR02.
   For the work-energy principle we have
      Wnet = DK + DU;
      0 = [(1/2m1v2 + 1/2m2v2 + 1/2Iw2) - 0] + m1g(h - 0) + m2g(0 - h);
      1/2m1v2 + 1/2m2v2 + 1/2(1/2MR02)(v/R0)2 = (m2 - m1)gh;
      1/2[m1 + m2 + 1/2M]v2 = (m2 - m1)gh;
      1/2[35.0 kg + 38.0 kg + 1/2(4.8 kg)]v2 = 
         (38.0 kg - 35.0 kg)(9.80 m/s2)(2.5 m), which gives
      v =        1.4 m/s.


75.   If the ball rolls without slipping, the speed of the center of 
   mass is v = r0w.  Because energy is conserved, for the motion 
   from  A to B  we have
      0 = DK + DU;
      0 = [(1/2mv2 + 1/2Iw2) - 0] + (mgr0 - mgR0),  or 
      1/2mv2 + 1/2(2/5 mr02)(v/r0)2 = mg(R0 - r0), which gives
      v = [10g(R0 - r0)/7]1/2.


18.   We find the initial and final angular velocities of the wheel from the rolling condition:
      w0 = v0/r = [(90.0 km/h)/(3.6 ks/h)]/(0.45 m) = 55.6 rad/s;
      w = v/r = [(60.0 km/h)/(3.6 ks/h)]/(0.45 m) = 37.0 rad/s.
   (a)   We find the angular acceleration from
         w2 = w02 + 2aq;
         (37.0 rad/s)2 = (55.6 rad/s)2 + 2a (85 rev)(2p rad/rev), which gives a=      - 1.6 rad/s2. 
   (b)   We find the additional time from
         wfinal = w + at;
         0 = 37.0 rad/s + (- 1.6 rad/s2)t,  which gives t =        23 s.


24.   (a)   We let L be the length of the beam and take clockwise as the 
      positive direction.  For the net torque about point C, we have
         tC    = 1/2L(F1 sin q1) - 1/2L(F3 sin q3) 
            = 1/2(2.0 m)(50 N) sin 30 - 1/2(2.0 m)(50 N) sin 60 
            =      - 18 m  N  (CCW=counter-clock-wise).
   (b)   For the net torque about point P, we have
         tP    = L(F1 sin q1) - 1/2L(F2 sin q2) 
            = (2.0 m)(50 N) sin 30 - 1/2(2.0 m)(60 N) sin 45 
            =      7.6 m  N  (CW=clockwise).


(See FIGURE 10-56)
31.   For the moment of inertia of the rotor blades we have
      I  = 3(1/3mbladeL2) = mbladeL2 = (160 kg)(3.75 m)2 =         2.25 x103kg  m2.
   We find the required torque from
      t    = Ia = I(w - w0)/t 
         = (2.25 x10-3kg  m2)[(5.0 rev/s)(2p rad/rev) - 0]/(8.0 s) =        8.8 x103m  N.


46.   (a)   If we treat the sphere as a point mass, we get
         Ia = MR02.
   (b)   We use the parallel-axis theorem:
         Ib  = ICM + Mh2 =       (2MR12/5) + MR02.
   (c)   The error is 
         error    = (Ia - Ib)/Ib  
               = [MR02 - (2MR12/5) - MR02]/[(2MR12/5) + MR02] 
               = [- (2R12/5)]/[(2R12/5) + R02]  
               = - (0.40)(0.10 m)2/[(0.40)(0.10 m)2 + (1.0 m)2] = - 0.0040 =      - 0.40%.


92.   (a)   The angular acceleration of the ball-arm system is
         a = atan/d1 = (7.0 m/s2)/(0.30 m) = 23.3 rad/s2. 
      Because we ignore the mass of the arm, for the moment of 
      inertia we have
         I = mballd12 = (1.00 kg)(0.30 m)2 = 0.090 kg  m2.
      Thus we find the required torque from
         t    = Ia 
            = (0.090 kg  m2)(23.3 rad/s2) =        2.1 m  N.
   (b)   Because the force from the triceps muscle is perpendicular 
      to the line from the axis, we find the force from
         F = t/d2 = (2.1 m  N)/(0.025 m) =       84 N.


105.   (a)   The speed of the block is the tangential speed 
      of the cylinder, so v = Rw.  For the system of 
      block and cylinder, if there is no friction, energy 
      is conserved.  After the block has moved a 
      distance D, we have
         0 = DK + DU;
         0 = [(1/2mv2 + 1/2I1w2) - 0] + (0 - mgD sin q); 
         1/2mv2 + 1/2(1/2MR2)(v/R)2 = mgD sin q; 
         1/2mv2 + 1/4Mv2 = mgD sin q;
         [1/2(3.0 kg) + 1/4(30 kg)]v2 = 
               (3.0 kg)(9.80 m/s2)(1.80 m) sin 30,
      which gives
         v =       1.7 m/s.
   (b)   From the force diagram for the block we see that
         Ffr2 = mFN2 = mmg cos q.
      For the cylinder, we assume a single vertical normal force.  Because there is no linear acceleration 
      of the center of mass, we have
         FN1 = Mg + FT sin q.
      Because the block accelerates down the plane, FT < mg sin D, and m = 0.1M.  Thus we ignore the 
      contribution of FT sin q (and we are justified in ignoring the horizontal component of FN1) to get
         FN1 ~ Mg, and Ffr1 = mFN1 = mMg.
      When the block moves a distance D, the surface of the rotating cylinder will move a distance D 
      through the depression.  For the work-energy theorem we have
         Wfr = DK + DU;
         - Ffr1D - Ffr2D = [(1/2mv2 + 1/2I1w2) - 0] + (0 - mgD sin q); 
         - m(M + m cos q)gD = 1/2mv2 + 1/4Mv2 - mgD sin q;
         - (0.035)[30 kg + (3.0 kg) cos 30](9.80 m/s2)(1.80 m) = 
                           [1/2(3.0 kg) + 1/4(30 kg)]v2 - (3.0 kg)(9.80 m/s2)(1.80 m) sin 30,
      which gives
         v =       0.84 m/s.


(See FIGURE 10-44)
79.   (a)   From Example 10-24, we see that while the ball is slipping, the acceleration of the center of mass 
      is - mkg and is constant.  The ball slips for a time T = 2v0/7mkg, so we find the distance from
         xslip = v0t + 1/2at2 
            = v0(2v0/7mkg) + 1/2(- mkg)(2v0/7mkg)2 =      12v02/49mkg.
   (b)   Once the ball starts rolling at time T, the linear speed is constant
         v = v0 + at = v0 + (- mkg)(2v0/7mkg) =       5v0/7.
      We can find the angular speed from
         w = w0 + at,  or from
         w = v/R = (5v0/7)/R =       5v0/7R