A cord is wound round the circumference of wheel of radius r.

Question:

A cord is wound round the circumference of wheel of radius $\mathbf{r}$. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight $\mathrm{mg}$ 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:

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

  2. 2gh

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

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


Correct Option: , 4

Solution:

using energy conservation between $\mathrm{A}$ and $\mathrm{B}$ point

$m g h=\frac{1}{2} m(w R)^{2}+\frac{1}{2} I \omega^{2}$

$2 \mathrm{mgh}=\left(\mathrm{MR}^{2}+\mathrm{I}\right) \omega^{2}$

$\omega^{2}=\frac{2 m g h}{I+M R^{2}}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now