A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance ‘h’, the square of angular velocity of wheel will be:

Option 1 -

$\frac{2 m g h}{I + m r^{2}}$

Option 2 -

$2 g h$

Option 3 -

$\frac{2 g h}{I + m r^{2}}$

Option 4 -

$\frac{2 m g h}{I + 2 m r^{2}}$

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

mg – T = ma. (1)

T × R = l . (2)

a = R. (3)

With the help of equations (1), (2) and (3), we get

$a = \frac{m g}{m + \frac{l}{R^{2}}}$

$v = \sqrt[]{2 a h} = ω R$

$\Rightarrow ω^{2} = \frac{2 m g h}{l + m R^{2}}$

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