A cylinder with moment of inertia 18.8 kg m2 rotates with angular velocity 7.85

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

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