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. velocity is \( v \), acceleration is zero
• B. velocity is zero, acceleration is zero
• C. velocity is \( v \), acceleration is \( \frac{v^2}{R} \)
• D. velocity is zero, acceleration is \( \frac{v^2}{R} \)

Hint:

In pure rolling motion, the velocity at the point of contact is zero, but the acceleration is non-zero and is given by \( \frac{v^2}{R} \).

Answer 1

The Correct Option is D

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: \[ \boxed{velocity = 0, \, acceleration = \frac{v^2}{R}} \]