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
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
Updated On: Jul 6, 2022
- $\frac {MR}{2T}$
- $\frac {2T}{MR}$
- $\frac {T}{MR}$
- $\frac {MR}{T}$
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The Correct Option is B
Solution and Explanation
The torque exerted on the disc is given, by
$\hspace15mm \tau=TR \hspace15mm ...(i)$
Also $\hspace15mm \tau=1 \alpha \hspace15mm ...(ii)$
From Eqs. (i) and (ii), we get
$\hspace15mm I\alpha=TR$
$\hspace15mm \alpha=\frac {TR}{I}= \frac {2TR}{MR^2}$
or $\hspace15mm \alpha=\frac {2T}{MR}$
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Concepts Used:
System of Particles and Rotational Motion
- The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
- The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
- The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.