Question 507: Centre of Mass and Rotational Motion

Question

A body slides down a smooth inclined plane of inclination $	heta$ and reaches the bottom with velocity V . If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of plane is

Accepted Answer

Let's solve step by step:
Step 1: Sliding Body (Smooth Incline, No Rotation)
The body descends a height $h$, starting from rest.
By law of conservation of energy:
$ 	ext{Potential Energy at Top} = 	ext{Kinetic Energy at Bottom} $
$ mgh = \frac{1}{2}mv^2 $
Solving for $v$:
$ v = \sqrt{2gh} $
You are given that this velocity is $V$.
$ V = \sqrt{2gh} $
Step 2: Rolling Ring (No Slipping, Both Translational and Rotational Kinetic Energy)
For a rolling ring:
Mass = $m$,
Velocity at bottom = $v$,
Moment of inertia about centre = $I = mR^2$,
Angular velocity $= \omega = \frac{v}{R}$.
Law of conservation of energy:
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\left(\frac{v}{R}ight)^2 $
$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}m v^2 $
$ mgh = mv^2 $
$ v^2 = gh $
$ v = \sqrt{gh} $
Step 3: Relate to $V$
$ V = \sqrt{2gh} \implies \sqrt{gh} = \frac{V}{\sqrt{2}} $
So, for the rolling ring, the linear velocity at the bottom is:
$ \boxed{\frac{V}{\sqrt{2}}} $
Correct Option: (A) $\frac{V}{\sqrt{2}}$

Answer

$\frac{\mathrm{V}}{\sqrt{2}}$

Answer

V

Answer

2 V

Answer

$\frac{\mathrm{V}}{2}$

Centre of Mass and Rotational Motion

A body slides down a smooth inclined plane of inclination \theta and reaches the bottom with velocity V . If the same body is a ring which rolls down the same inclined plane then linear velocity at the bottom of plane is

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