Rotational Motion Question 133
Question: A uniform disc of mass $M$ and radius $R$ is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc and a steady downward pull $T$ is exerted on the cord. The angular acceleration of the disc is
Options:
A) $\frac{T}{MR}$
B) $\frac{MR}{T}$
C) $\frac{2T}{MR}$
D) $\frac{MR}{2T}$
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Answer:
Correct Answer: C
Solution:
[c]Torque exerted on the disc $\tau =TR$
$ Now $ $\tau =I\alpha $ $\alpha =\frac{\tau }{I}=\frac{TR}{\frac{1}{2}MR^{2}}$ $=\frac{2TR}{MR^{2}}=\frac{2T}{MR}$
