A string is wrapped around a wheel of radius ‘r’. The axis of the wheel is horizontal and its M.I. about the axis is I. Weight ‘mg’ is tied to free end of the string which is released to fall down from rest position. The angular velocity of the wheel after it has fallen through distance ‘h’ will be

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

${(\frac{2 g h}{I + m r^{2}})}^{\frac{1}{2}}$

$\sqrt{2 g h}$

${(\frac{2 m g h}{I + m r^{2}})}^{\frac{1}{2}}$

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The change in potential energy of the weight will be converted into the kinetic energy of the wheel plus weight

$m g h = \frac{1}{2} m ϑ^{2} + \frac{1}{2} I ω^{2}$

= $\frac{1}{2} m r^{2} ω^{2} + \frac{1}{2} I ω^{2}$

$m g h = \frac{1}{2} ω^{2} (m r^{2} + I)$

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

Watch 3-min video & get full concept clarity