Question

A body slides down a smooth inclined plane of inclination \(θ\) 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 the plane is

• A. \(\frac{V}{\sqrt{2}}\)
• B. \(\frac{V}{2}\)
• C. \(V\)
• D. \(2V\)

Hint:

Compare energy conservation for sliding with energy conservation for rolling, where rotation takes half the energy for a ring.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
On a smooth plane the body slides, so all the lost potential energy becomes translational kinetic energy. A ring rolling without slipping shares the energy between translation and rotation.

Step 2: Sliding body.
\[ mgh = \frac{1}{2}mV^2 \Rightarrow V^2 = 2gh \]

Step 3: Rolling ring.
For a ring \(I = mR^2\) and \(\omega = v/R\). The kinetic energy is
\[ \frac{1}{2}mv^2 + \frac{1}{2}mR^2\left(\frac{v}{R}\right)^2 = mv^2 \]
Energy conservation gives \(mgh = mv^2\), so \(v^2 = gh\).

Step 4: Compare.
\(v^2 = \dfrac{V^2}{2}\), so \(v = \dfrac{V}{\sqrt{2}}\).

Step 5: Check the options.
A speed of \(V\) would mean no energy goes into rotation, and \(V/2\) or \(2V\) do not match the equal split.

Final Answer:
The ring reaches the bottom with speed \(\dfrac{V}{\sqrt{2}}\), option (A). \[ \boxed{\frac{V}{\sqrt{2}}} \]