ANSWER TO CHAPTER 11 HOMEWORK

1.  For the torque we have

                      t = r x F = [(4.0i + 8.0j + 6.0k)m] x [(16.0j - 4.0k)N]

                        = {[(8.0)(-4.0) - (6.0)(16.0)]i + [0 -(4.0)(-4.0)]j + [(4.0)(16.0) - 0]k}m.N

                        =  (-128i +16j + 64k)m.N.

2.  For   the angular momentum we have

                      l = r x mv = (mr x v)

                        = (0.060kg)[(7.0m)i + (-6.0m)j] x [(2.0m/s)i - (8.0m/s)k]

                        = (0.060kg){[(-6.0)(-8.0m/s) - 0]i + [0 -(7.0m)(-8.0m/s)]j + [0 - (-6.0m)(2.0m/s)]k}

                        =  (2.9i + 3.4j + 0.72k)kg.m2/s.

3.  (a)  With the positive direction CCW, for the angular momentum about the axis of the pulley we have

                       L = RoM1v + RoM2v + Iw

                          = RoM1v + RoM2v + I(v/Ro) =  [RoM1 + RoM2 + (I/Ro)v.

         (b)  Because M1g is balanced by the normal force on the horizontal surface, the net torque is from M2g only:

                       t = dL/dt;

                      M2gR = [RoM1 + RoM2  +(I/Ro)]dv/dt, which gives

                      a = M2g/[M1 + M2 + (I/Ro2)].

  

4.  For the system of stick and bullet during the collision, angular momentum about the center of mass is conserved:

                        Li = Lf.    

                        mvi(1/4l) + 0 = mvf(1/4l)  + Irodw;   

                       (3.0g)(250m/s)1/4(1.0m) = (3.0g)(160m/s)1/4(1.0m) + [(300g)(1.0m)2/12)]w

           which gives

                       w =  2.7rad/s.

*5 We select a differential element of the rod a distance r from the center, which has a mass dm  = (M/l)dr.

The angular momentum of this element is

                         dL = r x vdm,

Where v has magnitude (rsinf)w.

From the diagram, we see that for the lower half of the rod, r and v reverse direction. Thus the direction of dL will be the same for all the elements, and L will be perpendicular to the rod, making an angle  90° - f  with the axis.

We integrate to find the magnitude of the angular momentum:

                         L = ò dL  = ò  r(rsinf)wdm = ò -l/2l/2r(rsinf)w(M/l)dr

                            = (wM/l)sinf(r3/3)|-l/2l/2 

                            = (wM/l)sinf(2/3)(l/2)3 =  (1/12)Ml2wsinf.

 
*6. (a) We find the speed of the center of mass from the conservation of linear momentum:

                         Mv + 0 = (M + m)vCM;

                         (200kg)(18m/s) = ( 200kg + 50kg)vCM, which gives vCM = 14m/s.

  (b) During the collision, angular momentum about the center of mass willl be conserved. We find the location of the center of mass relative to the center of the beam:

                          d = m(1/2l)/(M + m) 

                             = (50kg)1/2(2.0m)/(200kg + 50kg)

                             = 0.20m.

When we use the parallel-axis theorem for the moment of inertia of the beam, angular momentum conservation gives us

                         Li = Lf;

                        Mvd + 0 = Itotalw = [(Ml2/12) + Md2 + m(1/2l - d)2]w;

                        (200kg)(18m/s)(0.20m) = ((200kg){[(2.0m)2/12] + (0.20m)2}+ (50kg)(1.0m - 0.20m)w;

which gives      w = 6.8rad/s.                     

 

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