Question

A hollow cylinder of mass $M$ and radius $R$ is pulled by a horizontal force $F$ acting at its centre of mass on a horizontal surface where the coefficient of friction is $\mu$. The maximum acceleration of the cylinder so that it may not start slipping is :

• A. $2\mu g$
• B. $\mu g$
• C. $\frac{2}{3}\mu g$
• D. $\frac{3}{2}\mu g$

Answer 1

The correct option is B $\mu g$

If slipping does not occur, point P is an instant center of rotation for the cylinder.

The equation of motion of the cylinder about point P is

${\tau }_P=I_P\alpha$ where

${\tau }_P$ is the torque acting about point P

$I_P$ is MI of the cylinder about point P.

$\alpha$ is the angular acceleration about point P.

$\therefore FR=(I_O+M{R}^2)\alpha$ .....(1) by parallel axis theorem, $I_P=I_O+M{R}^2$, $I_O$ is MI of the cylinder about cm O.

$\therefore \alpha =\frac{FR}{I_O+M{R}^2}$ ......(2)

Frictional force$F_{fr}$ is responsible for angular acceleration about cm O.

${\tau }_{fr}=F_{fr}R=I_O\alpha$ .....(3)

Substituting $\alpha$ from (2) in (3)

$F_{fr}R=I_O\frac{FR}{I_O+M{R}^2}$

$\Rightarrow F_{fr}=I_O\frac{F}{I_O+M{R}^2}$ .....(4)

As seen from (4) The applied force F and the frictional force$F_{fr}$ are proportional to each other. If F increases, $F_{fr}$ also increases until it reaches its possible max value.

This limiting value is given by $(F_{fr}{)}_{max}\mu N=\mu Mg$ .....(5)

Combining (4) and (5)

$\mu Mg=I_O\frac{F_{max}}{I_O+M{R}^2}$

$\Rightarrow F_{max}=\mu Mg(1+\frac{M{R}^2}{I_O})$ ......(6)

For a hollow cylinder $I_O=M{R}^2$

Substituting $I_O=M{R}^2$ in (6) we get $F_{max}=\mu Mg(1+\frac{M{R}^2}{M{R}^2})=2\mu Mg$

Using (2) ${\alpha }_{max}=\frac{F_{max}R}{I_O+M{R}^2}=\frac{2\mu MgR}{M{R}^2+M{R}^2}=\frac{\mu g}{R}$ .......(7)

$\therefore a_{max}={\alpha }_{max}R=\frac{\mu g}{R}\ast R=\mu g$