`ANSWER TO CHAPTER 10 HOMEWORK

1.  (a)  w = (2500rev.min)(2prad/rev))/(60s/min) = 262 rad/s.

         (b)  The linear speed of the point on the edge is the tangential speed:

                      v = rw = (0.175m)(262rad/s) = 46m/s.

                Because the speed is constant, the tangential acceleration is zero. There will be a radial acceleration:

                      aR = w2R = (262rad/s)2(0.175m) =  1.2 x 104m/s2 radial.

2.  (a)  We find the instantaneous angular velocity by differentiating:

                      w = dq/dt = d[(6.0rad/s)t - (8.0rad/s2)t2 + (4.5rad/s4)t4]/dt

                          = (6.0rad/s) - (16.0rad/s2)t + (18.0rad/s4)t3.

         (b)  We find the instantaneous angular acceleration by differentiating:

                       a =dw/dt = d[(6.0rad/s) - (16.0rad/s2)t + (18.0rad/s)4t3]/dt

                          = -(16.0rad/s2) + (54.0rad/s4)t2.

          (c)  At t = 3.0s, we have

                        w = (6.0rad/s) - (16.0rad/s2)(3.0s) + (18.0rad/s4)(3.0s)2 = 444 rad/s =   4.4 x 102rad/s;

                         a = -(16.0rad/s2) + (54.0rad/s4)(3.0s)2 = 470rad/s2 = 4.7 x 102rad/s2.

           (d)  The angular positions at the two times are

                          q2 = (6.0rad/s)(2.0s) - (8.0rad/s2)(2.0s)2 + (4.5rad/s4)(2.0s)4 = 52rad;

                          q3 = (6.0rad/s)(3.0s) - (8.0rad/s2)(3.0s)2 + (4.5rad/s4)(3.0s)4 = 311rad.

                  We find the average angular velocity from

                          wav = Dq/Dt = (311rad - 52rad)/(3.0s - 2.0s) =  2.6 x 102rad/s.

            (e)  The angular velocities at the two times are

                          w2 = (6.0rad/s) - (16.0rad/s2)(2.0s) + (18.0rad/s4)(2.0s)3 = 118rad/s;

                          w3 = 444rad/s from part (c).

                  We find the average angular acceleration from

                           aav = Dw/Dt = (444rad/s - 118rad/s)/(3.0s - 2.0s) =   3.3 x 102rad/s.

                  Note that the average angular velocity in part (d) is not 1/2(w2 + w3). The angular acceleration is not constant!

          

3.  (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(F1sinq1) - 1/2L(F3sinq3)

                                = 1/2(2.0m)(50N)sin30° - 1/2(2.0m)(50N)sin60°

                                = -18m/N (CCW).

          (b)  For the net torque about point P, we have

                            tP = L(F1sinq1) - 1/2L(F2sinq2)

                                 = (2.0)(50N)sin30° - 1/2(2.0m)(60N)sin45°

                                 =  7.6m.N (CW).

  

4. (a) Because we can ignore the mass of the rod, for the moment of inertia we hve

                              I = mballR2

                                = (2.4kg)(1.2m)2 =  3.5kg.m2.      

        (b)  To produce constant angular velocity, the net torque must be zero:

                              tnet = tapplied - tfriction = 0, or  

                              tapplied = FfrR = (0.020N)(1.2m) =   0.024m.N. 

5.  (a)  See the diagram at right.

        ( b)  We take the positive direction as the direction of the acceleration

                for each block and clockwise for the pulley.

                We apply SF = ma to each clock to find the tensions;

                        FT1 - m1gsinq1 = m1a;

                        FT1 - (8.0kg)(9.80m/s2)sin30°

                         = (8.0kg)(1.00m/s2),

                 which gives  FT1 = 47N.

                        m2gsinq2 - FT2 = m2a;

                        (10.0kg)(9.80m/s2)sin60° - FT2

                         = (10.0kg)(1.00m/s2)

               which gives FT2 = 75N.

         (c)  The net torque acting on the pulley is 

                         tnet = FT2R - FT1R = (75N)(0.25m) - (47N)(0.25m)

                           = 7.0m.N.

                We apply St = Ia to find the moment of inertia:

                          tnet = Ia = I(a/R)

                          7.0m.N = I(1.00m/s2)/(0.25m), 

                which gives I =  1.7kg.m2.

  
6.  (a)  We use the parallel-axis theorem:

                          I = 2(ICM +Mh2) = 2[(2MRo2/5 + M(3Ro/2)2] = 5.30MRo2.

          (b) If we treat the spheres as point masses, we get

                          I' = 2[M(3Ro/2)2] = 4.50MRo2.

                The error is

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

 

  
7.  For the system of the two blocks and pulley, no work will be done by non-conservative forces.

         The rope ensures that each block has the same speed v and the angular speed of the pulley is w = v/R.o.

         We choose the reference level for gravitational potential energy at the floor.

         The rotational inertia of the pulley is I = 1/2MRo2.

         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(1/2MRo2)(v/Ro)2 = (m2- m1)gh; 

                      1/2[m1 + m2 + 1/2M]v = (m2 - m1)gh;

                      1/2[35.0kg + 38.0kg + 1/2(4.8kg)]v2

                       = (38.0kg - 35.0kg)(9.80m/s2)(2.5m),   which gives

                       v =   1.4m/s

  
8.  (a)  If the pipe rolls without slipping, the speed of the center of mass is v = Rw.

               Because energy is conserved, from the top to the bottom of the incline we have

                         0 = DK + DU;

                         0 = [(1/2mv2 + 1/2Iw2) - 0] + (0 - mgDsinq), or

                         1/2mv2 + 1/2(mR2)(v/R)2 = mv2 = mgDsinq, which gives

                         v = (gDsinq)1/2 = [(9.80m/s2)(5.60m)sin21.5°]1/2 =  4.48m/s.

          (b)  The kinetic energy is

                         K = 1/2mv2 + 1/2Iw2 = 1/2mv2 + 1/2(mR2)(v/R)2

                             = mv2 = (0.0600kg)(4.48m/s)2 =  1.21J.

          (c)  For the rolling pipe, the acceleration around the center of mass and the angular acceleration 

                are related:  a = Ra. From the force diagram we have

                        FN = mgcosq, and mgsinq - Ffr = ma.

                For the angular acceleration around the center of mass, we have

                        FfrR = Ia = (mR2)(a/R), or Ffr = ma.

               When we use this in the force equation, we get

                       Ffr = 1/2mgsinq.

               If the pipe is not to slip, the friction force must be less than the maximum static friction:

                      Ffr <  msFN;

                      1/2mgsinq < msmgcosq;

                      1/2tan21.5° < ms, which gives  ms > 0.197.

  
*9. (a) Although there are forces at the point A at the tabletop, they create no torque about oint A. For the angular motion we have

                        tA = IAa;

                        mglcosf = 1/3ml2a = 1/3ml2(dw/dt), or

                        dw/dt = (3g/2l)cosf 

This equation contains three variables, but we can eliminate one from the definition of angular velocity: w = dq/dt = df /dt.

If we use each side as a multiplier of the previous equation, we get

                        wdw/dt = (3g/2l)(-f /dt)cosf , or wdw = - (3g/2l)cosf df .  

We integrate to get w(f):

                        ò 0w wdw =  -(3g/2l)ò p/2wcosf df;              

                        w2/2|0w = -(3g/2l)sinf|p/2w = -(3g/2l)(sinf - 1),

which gives w2 = (3g/l)(1 - sinf).

(b) At the tabletop, f = 0, so we have

                        w2 = (3g/l)(1 - 0) , or w = (3g/l)1/2.

The speed of the tip is

                        v= wl = l(3g/l)1/2 = (3g/l)1/2.                 

*10 (a) We select a differential element of the plate with sides dx and dy at the location x, y. The element is equivalent to a point mass with a mass of   dm = (M/lw)dxdy.

We integrate from x = -l/2 to x = l/2 and y = -w/2 to y = w/2 to find the moment of inertia of the plate:

                        I  = ò r2dm  = M/lwòò (x2 + y2)dxdy

                           = M/lw(ò-l/2l/2x2dxò-w/2w/2dy + ò-l/2l/2dxò-w/2w/2y2dy)

                           = M/lw(x3/3|-l/2l/2y|-w/2w/2 + x|-l/2l/2y3/3|-w/2w/2)

                           = M/3lw(2l3w/8 + 2lw3/8) = M/12(l2 + w2).

(b) For a axis along the edge parallel to the y-axis, we can consider the plate to be a infinite number of rods of length l and width dy with a mass (M/w)dy.

The moment of inertia of each rod is

                       dI = (l2/12)dm = (M/w)(l2/12)dx.

When we add (integrate), we get

                       I = (M/w)(l2/12)w = Ml2/12.

Similarly, for the other edge we get   I = Mw2/12.                      

                

 

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