Question

A rolling disc having linear acceleration $a$ and angular acceleration $\alpha$ is placed on a moving surface. The surface is moving with acceleration $a_s$, as shown in the image. The disc will be in pure rolling, when

• A. $a-R\alpha =a_s$
• B. $a+R\alpha =a_s$
• C. $\alpha -Ra=0$
• D. $\alpha +Ra=0$

Answer 1

The correct option is A $a-R\alpha =a_s$

For pure rolling there should not be any relative motion of the point of contact w.r.t. the surface on which the disc rolls.


From the figure,
Acceleration of contact point = Acceleration of surface
$\Rightarrow -R\alpha +a=a_s$
$\Rightarrow a-R\alpha =a_s$
Hence answer (a) is correct.