JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
One end of a straight uniform \(1\; \mathrm{m}\) long bar is pivoted on horizontal table. It is released from rest when it makes an angle \(30^{\circ}\) from the horizontal (see figure). Its angular speed when it hits the table is given as \(\sqrt{\mathrm{n}}\; \mathrm{s}^{-1},\) where \(\mathrm{n}\) is an integer. The value of \(n\) is
- A \(10\)
- B \(13\)
- C \(15\)
- D \(18\)
Answer & Solution
Correct Answer
(C) \(15\)
Step-by-step Solution
Detailed explanation
From mechanical energy conservation, \(\mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{r}}\) \(\Rightarrow \mathrm{mg} \frac{\ell}{2} \sin 30^{\circ}+0=0+\frac{1}{2} \mathrm{I} \omega^{2}\)…
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