Question

A disc of radius \( 2 m\) and mass \(100 kg\) rolls on a horizontal floor. Its centre of mass has speed of \(20 cm/s.\) How much work is needed to stop it? 

• A. \(3J\)
• B. \(30J\)
• C. \(2J\)
• D. \(1J\)

Answer 1

The Correct Option is A

To determine the work needed to stop the rolling disc, we must consider both its translational and rotational kinetic energy components. A disc rolling on a floor without slipping means it possesses both types of kinetic energy.

1. Translational Kinetic Energy: The kinetic energy due to the motion of the object's center of mass is given by: \( KE_{\text{trans}} = \frac{1}{2} m v^2 \)
   • Here, \( m = 100 \, \text{kg} \) is the mass of the disc, and
   • \( v = 20 \, \text{cm/s} = 0.2 \, \text{m/s} \) (converting from cm/s to m/s).
   • Substituting the values: \( KE_{\text{trans}} = \frac{1}{2} \times 100 \times (0.2)^2 = 2 \, \text{J} \)
2. Rotational Kinetic Energy: For a disc, the moment of inertia \( I \) around its center is given by: \( I = \frac{1}{2} m r^2 \), where \( r = 2 \, \text{m} \).
   • The angular velocity \\(\omega\\) is related to the linear velocity by \( \omega = \frac{v}{r} \).
   • Thus, \( \omega = \frac{0.2}{2} = 0.1 \, \text{rad/s} \).
   • The rotational kinetic energy is: \( KE_{\text{rot}} = \frac{1}{2} I \omega^2 \)
   • Substituting the values: \( I = \frac{1}{2} \times 100 \times (2)^2 = 200 \, \text{kg m}^2 \)
   • \( KE_{\text{rot}} = \frac{1}{2} \times 200 \times (0.1)^2 = 1 \, \text{J} \)
3. Total Kinetic Energy: The total kinetic energy of the rolling disc is the sum of the translational and rotational kinetic energies.
   • Total energy: \( KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 2 \, \text{J} + 1 \, \text{J} = 3 \, \text{J} \)

Thus, the total work required to stop the disc, which equates to the total kinetic energy it has, is \(3 \, \text{J}\).

The correct answer is, therefore, \(3J\).