Question

A cord is wound over the rim of a flywheel of mass 20kg and radius 25cm. A mass 2.5kg attached to the cord is allowed to fall under gravity. The angular acceleration of the flywheel is

Answer 1

A cord is wound over the rim of a flywheel of mass, M = 20 kg and radius, R = 25cm = 0.25m.

A mass,m = 2.5 kg is attached to the cord is allowed to fall under gravity.

here, torque = weight × perpendicular distance on weight from axis of rotation to circumference of rim as shown in figure.

or, $I\alpha=mg\times R$

here, I is moment of inertia of rim of flywheel, I = 1/2MR²

so, 1/2 MR² $\alpha$ = mgR

or, MR$\alpha$ = 2mg

or, $\alpha$ = 2mg/MR

= 2 × 2.5 × 10/20 × 0.25

= 10 rad/s²

hence, angular acceleration is 10 rad/s²

Answer 2

Answer:

$10 \frac{rad}{ {sec}^{2} }$

Explanation:

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