Question

A heavy homogenous cylinder has mass m and radius R. It is accelerated by a force F which is applied to the cylinder. The coefficient of static friction is sufficient for the cylinder to roll without slipping then

• A. The fractional force is $\frac{2F}{3}$
  
• B. The acceleration of the centre of the cylinder is $\frac{2F}{3mR}(R+r)$
  
• C. It is possible to choose 'r' so that acceleration of centre of cylinder is greater than $\frac{F}{m}$
  
• D. The direction of frictional force is always opposite to F

Answer 1

Answer: (B), (C)

The correct options are
B

The acceleration of the centre of the cylinder is $\frac{2F}{3mR}(R+r)$

C

It is possible to choose 'r' so that acceleration of centre of cylinder is greater than $\frac{F}{m}$

Let f be the frictional force: $F-f=m{a}_{cm}$ and $F_r+f_R=I\alpha =\frac{m{R}^2}{2}\alpha$

For pure rolling $\alpha R=a_{cm}$ Hence $\frac{F-f}{m}=\frac{2(Fr+fR}{mR}$ or FR-fR = 2Fr + 2fR

$f=\frac{F(-2r+R)}{3R}=\frac{2F}{3}(-\frac{r}{R}+\frac{1}{2})$ and $a_{cm}=\frac{F}{m}-\frac{2F}{3m}(-\frac{r}{R}+\frac{1}{2})=\frac{2f}{3mR}(\frac{3R}{2}+r-\frac{R}{2})=\frac{2F}{3mR}(R+r)$

If $a_{cm}>\frac{F}{m};\frac{2F}{3mR}(R+r)>\frac{F}{m}$ or $2R+2r>3R$ or $r>\frac{R}{2}$

It is possible that $a_m>\frac{F}{m}$ if $r>\frac{R}{2}$ or For $r=\frac{R}{2}$

$f=\frac{2R}{3}(\frac{1}{2}-\frac{1}{2})=0$ or if $r<\frac{R}{2},$ the friction force acts in the same directionas F