A long thin rod of mass M = 2.00 kg and length L = 75.0 cm is free to rotate abo
ID: 1402896 • 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 = .413 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.
(a) When they have moved halfway to the end of the rod, how fast (rad/s) is the rod rotating?
(b) 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)?
(c) When they reach the end, how fast is the rod rotating (rad/s)?
Explanation / Answer
let w1 and w2 are initial and final angular velocities
a) Apply Cosrvation of angular momentum
L1 = L2
I1*w1 = I2*w1
(M*L^2/12)*w1 = (M*L^2/12 + 2*m*(L/4)^2)*w2
w2 = (M*L^2/12)*w1/(M*L^2/12 + 2*m*(L/4)^2)
= (2*0.75^2/12)*10/(2*0.75^2/12 + 2*0.413*(0.75/4)^2)
= 7.64 rad/s
b) Kf/Ki = 0.5*I1*w1^2 / 0.5*I2*w2^2
= I1*w1^2/(I2*w2^2)
= (2*0.75^2/12)*10^2/((2*0.75^2/12 + 2*0.413*(0.75/4)^2 )*7.64^2)
= 1.308
c) Again Apply Cosrvation of angular momentum
L1 = L2
I1*w1 = I2*w1
(M*L^2/12)*w1 = (M*L^2/12 + 2*m*(L/4)^2)*w2
w2 = (M*L^2/12)*w1/(M*L^2/12 + 2*m*(L/2)^2)
= (2*0.75^2/12)*10/(2*0.75^2/12 + 2*0.413*(0.75/2)^2)
= 4.47 rad/s