JEE MAIN - Physics (2024 - 8th April Evening Shift - No. 26)
A circular table is rotating with an angular velocity of $$\omega \mathrm{~rad} / \mathrm{s}$$ about its axis (see figure). There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of $$1 \mathrm{~m}$$ on the groove. All the surfaces are smooth. If the radius of the table is $$3 \mathrm{~m}$$, the radial velocity of the ball w.r.t. the table at the time ball leaves the table is $$x \sqrt{2} \omega \mathrm{~m} / \mathrm{s}$$, where the value of $$x$$ is _________.
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Answer
2
Explanation
$$\begin{aligned}
& m v_r \frac{d v_r}{d r}=m \omega^2 r \\
& \frac{v_v^2}{2}=\frac{\omega^2\left(r^2-1\right)}{2} \\
& v_r=\omega \sqrt{8}=2 \omega \sqrt{2}
\end{aligned}$$
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