Question

A disc of mass 5 kg and radius 50 cm rolls on the ground at 10 m/s. Find the kinetic energy.

Hint:

For a rolling disc: Total KE = \( \frac{3}{4}mv^2 \) — memorize this shortcut!

Answer 1

Correct Answer: 375

Step 1: Understanding the Concept:
A rolling body has two components of kinetic energy: translational kinetic energy due to the motion of its center of mass and rotational kinetic energy due to rotation about its center of mass.
Step 2: Key Formula or Approach:
Total Kinetic Energy \(K = K_{trans} + K_{rot}\).
\[ K = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \]
For a disc, moment of inertia \(I = \frac{1}{2}mr^{2}\).
For pure rolling, \(\omega = \frac{v}{r}\).
Step 3: Detailed Explanation:
Given:
Mass \(m = 5\text{ kg}\).
Radius \(r = 50\text{ cm} = 0.5\text{ m}\).
Velocity \(v = 10\text{ ms}^{-1}\).
Substitute \(I\) and \(\omega\) into the total kinetic energy formula:
\[ K = \frac{1}{2}mv^{2} + \frac{1}{2} \left( \frac{1}{2}mr^{2} \right) \left( \frac{v}{r} \right)^{2} \]
\[ K = \frac{1}{2}mv^{2} + \frac{1}{4}mv^{2} = \frac{3}{4}mv^{2} \]
Insert the numerical values:
\[ K = \frac{3}{4} \times 5 \times (10)^{2} \]
\[ K = \frac{3}{4} \times 5 \times 100 \]
\[ K = 3 \times 5 \times 25 = 375\text{ J} \]
Step 4: Final Answer:
The total kinetic energy of the disc is \(375\text{ J}\).