A pulley, with a rotational inertia of3.2×10 -3 kg m 2 about its axleand a radiu

Question

A pulley, with a rotational inertia of3.2×10-3kg m2 about its axleand a radius of 6.0 cm, is acted on by a force appliedtangentially at its rim. The force magnitude varies in time as F =0.58t + 0.24t2, where F is in newtons if t is given inseconds. The pulley is initially at rest. At t = 2.6 swhat is its angular acceleration?

Explanation / Answer

We know that           = I         F * r = I ==> = F r / I             = [0.58t + 0.24t2] * 0.06 / 3.2×10-3             = [10.875 t + 4.5t2] at t = 2.6 s We have                 = 10.875 * 2.6 + 4.5 * 2.6 * 2.6 = 58.695 rad /s2

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