A long thin rod of mass M = 2.00 kg and length L = 75.0 cm is free to rotate abo
ID: 2052722 • Letter: A
Question
A long thin rod of mass M = 2.00 kg and length L = 75.0 cm is free to rotate about its center as shown. Two identical masses (each of mass m = .466 kg) slide without friction along the rod. The two masses begin at the rod's point of rotation when the rod is rotating at 10.0 rad/s. When they have moved halfway to the end of the rod, how fast (rad/s) is the rod rotating? When the masses are halfway to the end of the rod, what is the ratio of the final kinetic energy to the initial kinetic energy (Kf/ki)? When they reach the end, how fast is the rod rotating (rad/s)?Explanation / Answer
The length of the rod is L=.75m and mass M = 2kg
m = .466 kg.
initial angular velocity = 1= 10 rad/s
from the principle of conservation of angular momentum we have
I1*1 =I2*2.
here I1= initial moment of inertia of the system = M*L^2/12.
I2 = final moment of inertia of the system = M*L^2/12 + 2*m*r1^2.
where r= L/4
=> substituting the values we have
2 = 1*[ M/(M+3m/2)] = 7.41 rad/s.
b) initial kinetic energy K1 = 1/2 *I1*1^2.
final kinetic energy K2 = 1/2 *I2*2^2
=> K2/K1 = Kf/Ki = 2/1 = 7.41/10 = .741.
c) whenthe masses have reached the ends of the rod the moment of inertia of the rod is given by
I3 = M*L^2/12 + 2m*L^2/4
=> from the principle of conservation of angular momentum we have
I1*1 =I3*3.
=> we have 3 = I1*1/I3 = 4.17 rad/s