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 value of $F$ so that the cylinder may not slip is :

• A. $2\mu Mg$
• B. $3\mu Mg$
• C. $\frac{3}{2}\mu Mg$
• D. $\frac{5}{2}\mu Mg$

Answer 1

The correct option is A $2\mu Mg$

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 center of mass 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 $F_{max}=\mu Mg(1+\frac{M{R}^2}{M{R}^2})=2\mu Mg$