A string is wrapped around a pulley with a radius 2.0 cm. The pulley initially i

Question

A string is wrapped around a pulley with a radius 2.0 cm. The pulley initially is at rest. A constant force of 50 N is applied to the string, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2m in 4.9 s, what it the moment of inertia of the pulley.

A)14 kg x m^2
B)0.017 kg x m^2
C)17 kg x m^2
D)0.17 kg x m^2
E)0.20 kg x m^2

Please show step by step

Explanation / Answer

By v = u + at =>v = 0 + 4.9a =>v = 4.9a ------------(i) By v^2 = u^2 + 2as =>v^2 = 0 + 2 x v/4.9 x 1.2 =>4.9v^2 - 2.4v = 0 =>v(4.9v - 2.4) = 0 =>v = 2.4/4.9 = 0.49 m/s Thus by v = r x omega =>omega = v/r = 0.49/0.02 = 24.49 rad/sec BY W = F x s = 50 x 1.2 = 60 J =>KE(rotational) = W = 1/2 x I x omega^2 =>60 = 1/2 x I x (24.49)^2 =>I = 0.20 kg-m^2

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