JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A cord is wound round the circumference of wheel of radius \(r\). The axis of the wheel is horizontal and moment of inertia about it is \(I\). A weight \(mg\) is attached to the end of the cord and falls from rest. After falling through a distance \(h\), the angular velocity of the wheel will be
- A \(\sqrt {\frac{{2gh}}{{I + mr}}} \)
- B \(\sqrt {\frac{{2mgh}}{{I + m{r^2}}}} \)
- C \(\sqrt {\frac{{2mgh}}{{I + 2m{r^2}}}} \)
- D \(\sqrt {2gh} \)
Answer & Solution
Correct Answer
(B) \(\sqrt {\frac{{2mgh}}{{I + m{r^2}}}} \)
Step-by-step Solution
Detailed explanation
According to law of conservation of energy \(mgh = \frac{1}{2}(I + m{r^2}){\omega ^2}\) \( \Rightarrow \) \(\omega = \sqrt {\frac{{2mgh}}{{I + m{r^2}}}} \).
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