Question

A body is rolling down an inclined plane without slipping. How does the acceleration of the rolling body depend on its radius?

Answer 1

Step1: Given data

A body is rolling down an inclined plane without slipping.

Step 2: Formula used

$a=\frac{g\sin \theta }{\left(1+{\displaystyle \frac{I}{M{R}^2}}\right)}[wherea=acceleration,g=accelerationduetogravity,I=momentofinertia,M=mass,R=radius]$

Step 3: Relation between acceleration and radius

When a frictional force acts between the body and the surface, the body will roll. The torque required for rolling will be provided by this frictional force. When there is no frictional force, the body will slip from the inclined plane due to its own weight

$I=M{K}^2[k=radiusofgyration]$

Substituting the value of I in the acceleration formula

$$\begin{gathered} a=\frac{g\sin \theta }{1+{\displaystyle \frac{M{K}^2}{M{R}^2}}} \\ a=\frac{g\sin \theta }{1+{\displaystyle \frac{K^2}{R^2}}} \\ a\propto \frac{1}{1+{\displaystyle \frac{K^2}{R^2}}} \end{gathered}$$

Hence, the acceleration of the rolling body is directly proportional to its radius.