Home Physics System of Particles Rotational Motion JEE Main 2023 A light rope is wound around a hollow cylind…
Physics System of Particles Rotational Motion JEE Main 2023 MCQ (Single Correct)

A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be_____rad s -2 .

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The formula to calculate the moment of inertia (I) of the cylinder about its radial axis is given by

I = Mr 2 ...

Also, the torque (r) on the cylinder about the central axis because of the application of the external force is given by

= Fr ...

Also, the torque can be expressed as

= I ...

Substitute the expressions from equation and into equation and simplify to obtain the angular acceleration.

Fr= Mr 2

...

Substitute the values of the known parameters into equation to calculate the required angular acceleration. a = 52 - 5N

= 15 rad s -2

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