Question:
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in the figure. At any instant, for the lower most point of the disc
A disc is performing pure rolling on a smooth stationary surface with constant angular velocity as shown in the figure. At any instant, for the lower most point of the disc
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In pure rolling motion, the velocity at the point of contact is zero, but the acceleration is non-zero and is given by \( \frac{v^2}{R} \).
Updated On: Mar 24, 2026
- velocity is \( v \), acceleration is zero
- velocity is zero, acceleration is zero
- velocity is \( v \), acceleration is \( \frac{v^2}{R} \)
- velocity is zero, acceleration is \( \frac{v^2}{R} \)
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The Correct Option is D
Solution and Explanation
Step 1: Understand rolling motion.
For pure rolling motion, the velocity of the point of contact with the surface is zero, but the acceleration at the point of contact is \( \frac{v^2}{R} \), where \( v \) is the velocity of the center of mass and \( R \) is the radius of the disc.
Step 2: Conclusion.
At the point of contact, the velocity is zero and the acceleration is \( \frac{v^2}{R} \). Final Answer: \[ \boxed{velocity = 0, \, acceleration = \frac{v^2}{R}} \]
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