A cylinder with moment of inertia 18.8 kg m2 rotates with angular velocity 7.85
ID: 2265231 • Letter: A
Question
A cylinder with moment of inertia 18.8 kg m2 rotates with angular velocity 7.85 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 23.3 kgm2, initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity. final angular velocity is 2.97 rad/s.
Show that energy is lost in this situation by
calculating the ratio of the final to the initial
kinetic energy.
Explanation / Answer
I1*w1 = I2*w1
18.8*7.85 = (18.8+23.3)*w2
w2 = (18.8*7.85)/(18.8+23.3)
w2 = 3.5 rad/s
Ki = 0.5*I1*w1^2 = 0.5*18.8*7.85^2 = 579.2515 J
kf = 0.5*I2*w2^2 = 0.5*(18.8+23.3)*3.5^2 = 257.8625 J
kf<ki
so kinetic enrgy is lost