Physics · Horizontal Circular Motion

JEE Main 2024 — 8 April, Shift 2 — Question 57

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 m on the groove. All the surface 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 $\mathrm{x} \sqrt{2} \omega \mathrm{~m} / \mathrm{s}$ , where the value of x is $\qquad$

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$a_{c}=\omega^{2} x$

$\frac{v d v}{d x}=\omega^{2} x$

$\int_{0}^{v} v d v=\int_{1}^{3} \omega^{2} x d x$

$\frac{v^{2}}{2}=\omega^{2}\left[\frac{x^{2}}{2}\right]$

$\frac{v^{2}}{2}=\frac{\omega^{2}}{2}\left[3^{2}-1^{2}\right]$ $\mathrm{v}=2 \sqrt{2} \omega$

$\mathrm{x}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Horizontal Circular Motion
Topic: Dynamics of circular motion