Question

A heavy homogenous cylinder has a mass m and radius R. It is accelerated by a horizontal force F, which is applied through a rope wound around a light drum of radius r attached to the cylinder. The coefficient of static friction is sufficient for the cylinder to roll without slipping :
Choose the correct options :

• A. acceleration of centre of cylinder is $a=\frac{2F}{3mR}(R+r)$
• B. friction assumed to be in the opposite direction of F is $f=\frac{2}{3}[\frac{1}{2}-\frac{r}{R}]F$
• C. If $\left[\frac{r}{R}\right]<\frac{1}{2}$a will be greater than $\frac{F}{m}$
• D. Friction assumed to be in the direction of F is $f=\frac{2}{3}[\frac{1}{2}-\frac{r}{R}]F$

Answer 1

Answer: (A), (B)

If the acceleration of the system is $a$, then-

$F-f=ma$ ...........(1)

Now,

$Fr+fR=I\alpha$

$Fr+fR=\frac{m{R}^2}{2}\alpha$.

Since there is no slipping, $a=R\alpha$

$Fr+fR=\frac{mR}{2}a$

$\frac{2Fr}{R}+2f=ma$ ..........(2)

From equation 1 and 2

$\frac{2Fr}{R}+2f=F-f$

$\frac{2Fr}{R}-F=-3f$

$f=\frac{F}{3}\left(1-\frac{2r}{R}\right)$

Putting this in equation (i), we have-

$a=\frac{2F}{3m}\left(\frac{r+R}{R}\right)$

Now if,

$a>\frac{F}{m}$

$\frac{2F}{3m}\left(\frac{r+R}{R}\right)>\frac{F}{m}$

$\frac{2}{3}\left(\frac{r+R}{R}\right)>1$

$2r+2R>3R$

$2r>R$

$\frac{r}{R}>\frac{1}{2}$

Clearly, the direction of the friction will be forward.

So, the correct options are (A) & (B)