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3-2.Motion in Plane
hard
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 surface 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............
A
$1$
B
$2$
C
$5$
D
$7$
(JEE MAIN-2024)
Solution
$a_c=\omega^2 x$
$\frac{v d v}{d x}=\omega^2 x$
$\int_0^v v d v=\int_1^3 \omega^2 x d x$
$\frac{v^2}{2}=\omega^2\left[\frac{x^2}{2}\right]$
$\frac{v^2}{2}=\frac{\omega^2}{2}\left[3^2-1^2\right]$
$v=2 \sqrt{2} \omega$
$x=2$
Standard 11
Physics
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