Question

If a ring of mass 50 g and radius 2cm is rolling on a smooth horizontal platform with its centre of mass moving with a speed of 50 cms\(^{-1}\), then its total energy is

• A. \(1.0 \times 10^{-2}\) J
• B. \(1.25 \times 10^{-2}\) J
• C. \(2.5 \times 10^{-2}\) J
• D. \(3.5 \times 10^{-2}\) J
• E. \(1.5 \times 10^{-2}\) J

Hint:

For a rolling ring, the translational and rotational kinetic energies are equal.
Just calculate \(mv^2\) to get the total energy instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a body rolling without slipping, the total kinetic energy is the sum of translational kinetic energy and rotational kinetic energy.

Step 2: Key Formula or Approach:
\[ KE_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]
For a ring, the moment of inertia \(I = mR^2\) and \(\omega = v/R\).

Step 3: Detailed Explanation:
Substitute \(I\) and \(\omega\) into the total energy formula:
\[ KE_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} (mR^2) \left(\frac{v}{R}\right)^2 \]
\[ KE_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \]
Given values in SI units:
Mass \(m = 50 \text{ g} = 0.05 \text{ kg}\)
Velocity \(v = 50 \text{ cm/s} = 0.5 \text{ m/s}\)
\[ KE_{\text{total}} = 0.05 \times (0.5)^2 \]
\[ KE_{\text{total}} = 0.05 \times 0.25 = 0.0125 \text{ J} \]
This can be written as \(1.25 \times 10^{-2} \text{ J}\).

Step 4: Final Answer:
The total energy is \(1.25 \times 10^{-2} \text{ J}\).