# Which Cylinder is Faster?

## Objective

To see the effect of moment of inertia.

## Introduction

This experiment is based on IIT JEE 2012 question on rotational mechanics. There are two cylinders of equal mass and same outer dimensions. The mass of one of the cylinder, say Q, is concentrated near its axis and mass of other cylinder, say P, is concentrated near its periphery. Both the cylinder starts motion from the same height on an inclined plane. The friction is enough for rolling without slipping. Which cylinder reach the bottom faster and why?

## Procedure

There are many ways to make these cylinders. Take two pipes (of PVC or hard paper) of 1 inch diameter and 6 inch length. Take bicycle spokes of 12 inch length and cut them in two equal halves. Arrange the spokes on inner periphery of cylinder P such that mass is symmetrically distributed. You can either use adhesive like fevicol or put roll of paper inside the spokes for tight fitting. On another cylinder, put same number of spokes at the cylinder axis. Put paper roll above these spokes for tight fitting. Adjust number of spokes so that mass of both the cylinders is same. Your cylinders are ready.

## Discussion

Since cylinder P has most of its mass concentrated near its surface, its moment of inertia (about the cylinder axis) is more than the moment of inertia of the cylinder Q i.e., \(I_P>I_Q\). The forces acting on the cylinder are its weight \(mg\), normal reaction \(N\), and frictional force \(f\) (see figure). In case of rolling without slipping, \(v=\omega r\) and \(a=\alpha r\). The torque about centre of mass is related to \(\alpha\) by,

\begin{align} %\label{wca:eqn:1} \tau=rf=I\alpha, \nonumber \end{align}and the force along the plane is related to \(a\) by,

\begin{align} %\label{wca:eqn:2} mg\sin\theta-f=ma. \nonumber \end{align}Solve above equations to get,

\begin{align} %\label{wca:eqn:3} a=\frac{g\sin\theta}{1+I/(mr^2)}. \nonumber \end{align}
This equation gives \(a_P