A cylinder with moment of inertia I 1 rotates with angular velocity ? 0 about a

Question

A cylinder with moment of inertia I1 rotates with angular velocity ?0 about a frictionless vertical axle. A second cylinder, with moment of inertia I2, initially not rotating, drops onto the first cylinder (Fig. P8.55). Because the surfaces are rough, the two eventually reach the same angular velocity, ?.

Show that energy is lost in this situation (Do this on paper. Your instructor may ask you to turn in this work.), and calculate the ratio of the final to the initial kinetic energy. (Use I1 for I1, I2 for I2, w for ?, and w0 for ?0 in your equation.)

Explanation / Answer

L1 = L2 => (mVR)1 = (MVR)2 => v = r ?, I(cyl) = 0.5mr^2

I1 ?0 = (I1 + I2) ?, => ? = ?0 [I1/(I1 + I2)]

b)
E2/E1 = (I1 + I2) ?^2 / [I1 ?0^2]

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