SPRING 2002QUIZ 5

PHYSICS 220A #85077

1.Calculate the moment of inertia of a thin uniform bar about the axis through the point at 1/3L to one end and perpendicular to the bar as shown.  The bar has mass M and length L.

Solution.

 

 We choose the coordinate system as shown with the origin at the axis. we choose an infinitesimal length dx as mass element. we have

                  dm = (M/L)dx.

We calculate the moment of inertia about the axis

                 I = ò x2dm 

                    = ò -1/3L2/3L x2(M/L)dx

                    = (M/L)(1/3x3)|-1/3L2/3L

                    = (1/9)ML2.

2. A force F = (12.0i + 5.0j)N acts on a particle located at r = (4.5i - 7.0j + 6.0k)m. Calculate the torque about the origin due to the force.

Solution.

We find the torque from

                   t = r  x F

                      = (4.5i - 7.0j + 6.0k)m x (12.0i + 5.0j)N

                      = [0 - (6.0)(5.0)]i + [(6.0)(12.0) - 0]j +  [(4.5)(5.0) - (-7.0)(12.0)]k

                      = (-30.0i + 72.0j + 106.5k )m·N.