A wheel is rotating freely with an angular speed of 800 rev/minona shaft whose r

Question

A wheel is rotating freely with an angular speed of 800 rev/minona shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. What is the angular speed of the resultant combination of the shaft and two wheels ? What fraction of the original rotational kinetic energy is lost?

Explanation / Answer

Given that w1 = 800 rev/min I2 = 2 I1 a) From conservation of angular momentum we have I1 w1 = (I1 + I2) w2 ==> w2 = I1 w1 / (I1 + I2) = 800 * I1 / 3I1 = 266.66 rev/min b) Lost in KE = (Ki - Kf ) / Ki = (0.5 I1w1^2 - 0.5(I1+ I2) w2^2 ) / 0.5 I1 w1^2 = ( I1w1^2 -3 I1 w2^2 ) / I1 w1 = (w1^2 - 3w2^2) / w1^2 = (800^2 - 3 * 266.66^2) / 800^2 = 0.6666 = 2 / 3

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