Question

A circular turn table of radius 0.5 m has a smooth groove as shown in figure. A ball of mass 90 g is placed inside the groove along with a spring of spring constant l02 N/cm. The ball is at a distance of 0.1 m from the centre when the turn table is at rest. On rotating the turn .able with a constant angular frequency of 102 sec-1, the ball moves away from the centre by a distance nearly equal to.

( IIT Advance )

Rahul sharam !!​

Answer 1

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\boxed{10^{-2}\: m}$

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EXPLAINATION ➣

Since the disc is rotating in angular speed ω.$\green{\texttt{therefore particle(ball) must }}$$\green{\texttt{experience centrifugal force}}$.

But the ball is in rest so it is $\green{\texttt{in equilibrium}}$ and the spring force balancing the centrifugal force.

$Kx = mr {w}^{2}$

Given K = 10² N/cm = 10⁴ N/m

$= > {10}^{4} \times x = \frac{90}{1000} \times ( \frac{1}{10} ) \times ( {10}^{2} )^{2} \\ \\ = > {10}^{4} \times x = 90 \\ \\ = > 0.009 \\ \\ = > 0.01 \: approx = {10}^{ - 2} m$

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