Question:
A solid circular disc of mass 100 kg rolls along a horizontal floor so that center of mass has a speed of 0.2 m s\(^{-1}\). The absolute value of work done on the disc to stop it is.
A solid circular disc of mass 100 kg rolls along a horizontal floor so that center of mass has a speed of 0.2 m s\(^{-1}\). The absolute value of work done on the disc to stop it is.
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In rolling motion, total kinetic energy includes both translational and rotational parts.
Updated On: Jun 20, 2026
- 2 J
- 3 J
- 2.5 J
- 4 J
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The Correct Option is B
Solution and Explanation
Step 1: Understand rolling motion energy.
For a rolling solid disc: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For solid disc: \[ I = \frac{1}{2}mr^2, \quad v = \omega r \]
Step 2: Substitute rotational energy.
\[ K = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} \] \[ K = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 \] \[ K = \frac{3}{4}mv^2 \]
Step 3: Substitute values.
\[ K = \frac{3}{4} \times 100 \times (0.2)^2 \] \[ K = \frac{3}{4} \times 100 \times 0.04 \] \[ K = \frac{3}{4} \times 4 \] \[ K = 3 \, \text{J} \]
Step 4: Work done to stop the disc.
Work required = initial kinetic energy: \[ W = 3 \, \text{J} \]
Step 5: Final conclusion.
Thus, absolute work done to stop the rolling disc is: \[ \boxed{3 \, \text{J}} \]
For a rolling solid disc: \[ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For solid disc: \[ I = \frac{1}{2}mr^2, \quad v = \omega r \]
Step 2: Substitute rotational energy.
\[ K = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} \] \[ K = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 \] \[ K = \frac{3}{4}mv^2 \]
Step 3: Substitute values.
\[ K = \frac{3}{4} \times 100 \times (0.2)^2 \] \[ K = \frac{3}{4} \times 100 \times 0.04 \] \[ K = \frac{3}{4} \times 4 \] \[ K = 3 \, \text{J} \]
Step 4: Work done to stop the disc.
Work required = initial kinetic energy: \[ W = 3 \, \text{J} \]
Step 5: Final conclusion.
Thus, absolute work done to stop the rolling disc is: \[ \boxed{3 \, \text{J}} \]
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