A circular table is rotating with an angular velocity of ω rad/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 m on the groove. All the surfaces are smooth. If the radius of the table is 3 m, the radial velocity of the ball w.r.t. the table at the time ball leaves the table is x√(2) ω m/s, where the value of x is _________. Numerical answer type. Work it out, then open the answer.

Answer: 2. Table turns at constant ω; in its frame the ball feels only the centrifugal force mω²r along the groove. vdv/dr=ω²r⇒ v²=ω²(r²-r₀²) v²=ω²(3²-1²)=8ω²⇒ v=2√2 ω So x=2.

A circular table is rotating with an angular velocity of ω rad/s… | JEE Main 2024 (8 Apr, Shift 2)

Q27JEE Main 2024 · 8 APR · SHIFT 2Medium

Angular Momentum & its Conservation

A circular table is rotating with an angular velocity of $\omega\text{ rad/s}$ 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 $1\text{ m}$ on the groove. All the surfaces are smooth. If the radius of the table is $3\text{ m}$ , the radial velocity of the ball w.r.t. the table at the time ball leaves the table is $x\sqrt{2}\,\omega\text{ m/s}$ , where the value of $x$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Table turns at constant $\omega$ ; in its frame the ball feels only the centrifugal force $m\omega^2r$ along the groove.
$v\frac{dv}{dr}=\omega^2r\Rightarrow v^2=\omega^2(r^2-r_0^2)$
$v^2=\omega^2(3^2-1^2)=8\omega^2\Rightarrow v=2\sqrt2\,\omega$
So $x=2$ .

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