A thin, light wire is wrapped around the rim of a wheel, as shown in the followi

Question

A thin, light wire is wrapped around the rim of a wheel, as shown in the following figure. The wheel rotates without friction about a stationary horizontal axis that passes through the center of the wheel. The wheel is a uniform disk with radius 0.298 m . An object of mass 4.00 kg is suspended from the free end of the wire. The system is released from rest and the suspended object descends with constant acceleration.(Figure 1)

Question: If the suspended object moves downward a distance of 2.60 m in 1.99 s , what is the mass of the wheel?

Explanation / Answer

acceleration a = 2d / t² = 2 * 2.6m / (1.99s)² = 1.31 m/s²
tension T = m(g - a) = 4 kg * (9.8 - 1.31)m/s² = 33.96 N
and torque = T*r = 33.96N * 0.298m = 10.12 N·m
Also
= I = ½Mr² * a/r = ½Mra = ½M * 0.298m * 1.31m/s² = M * 0.39 m²/s²,
so 10.12 N·m = M * 0.39 m²/s²
M = 25.95 kg

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