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 disc, we need to calculate the total kinetic energy of the rolling disc, which includes both translational and rotational kinetic energy.

1. Translational Kinetic Energy: The formula for translational kinetic energy is given by: K_{\text{trans}} = \frac{1}{2} mv^2, where m is the mass and v is the velocity.
   • Here, m = 100 \, \text{kg} and v = 20 \, \text{cm/s} = 0.2 \, \text{m/s}.
   • Substituting the values, we get: K_{\text{trans}} = \frac{1}{2} \times 100 \, \text{kg} \times (0.2 \, \text{m/s})^2 = 2 \, \text{J}
2. Rotational Kinetic Energy: The formula for rotational kinetic energy is given by: K_{\text{rot}} = \frac{1}{2} I \omega^2, where I is the moment of inertia and \omega is the angular velocity.
   • For a disc, the moment of inertia is I = \frac{1}{2} m r^2.
     • Given r = 2 \, \text{m}, we calculate I as: I = \frac{1}{2} \times 100 \, \text{kg} \times (2 \, \text{m})^2 = 200 \, \text{kg} \cdot \text{m}^2.
   • The angular velocity \omega is related to the linear velocity by \omega = \frac{v}{r}. Therefore, \omega = \frac{0.2 \, \text{m/s}}{2 \, \text{m}} = 0.1 \, \text{rad/s}.
   • Substituting these values into the formula for rotational kinetic energy, we get: K_{\text{rot}} = \frac{1}{2} \times 200 \, \text{kg} \cdot \text{m}^2 \times (0.1 \, \text{rad/s})^2 = 1 \, \text{J}.
3. Total Kinetic Energy: The total kinetic energy of the rolling disc is the sum of its translational and rotational kinetic energies:
   • K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = 2 \, \text{J} + 1 \, \text{J} = 3 \, \text{J}
4. Since work done is equal to the change in kinetic energy, the work needed to stop the disc is equal to the total kinetic energy:
   • W_{\text{needed}} = 3 \, \text{J}

Therefore, the correct answer is 3 \, \text{J}.