Question

A uniform solid cylinder of mass m and radius 2R rests on a horizontal table. A string attached to it passes over a pulley (disc) of mass m and radius R that is mounted on a frictionless axle through its centre. A block of mass m is suspended from the free end of the string. The string does not slip over the pulley surface, and the cylinder rolls without slipping on the table. Force of friction acting on the cylinder is

• A. $\frac{2mg}{3}$
• B. $\frac{3mg}{2}$
• C. $\frac{mg}{3}$
• D. $\frac{mg}{6}$

Answer 1

The correct option is D $\frac{mg}{6}$

$Given$ :- Mass of cylinder = $m$

Radius of cylinder = $2R$

Mass of pulley = $m$

Radius of pulley = $R$

Mass of block = $m$

$ToFind$ :- Frictional force ( $f$ ) acting on cylinder

$Solution$ :- According to diagram ,

On Balancing force for block we get ,

$mg-{\tau }_1=ma$

${\tau }_1=mg-ma$ $-------\left(1\right)$

On Balancing force for cylinder we get ,

${\tau }_2-f=ma$

${\tau }_2=ma+f$ $-------\left(2\right)$

On Balancing force for pulley we get ,

$({\tau }_1-{\tau }_2)R=I_p{\alpha }_p$ $(let{\tau }_1>{\tau }_2)$

From $\left(1\right)and\left(2\right)$ we get ,

$(mg-2ma-f)R=\frac{m{R}^2}{2}{\alpha }_p$ $(\because I_p=\frac{m{R}^2}{2})$

$\therefore$ ${\alpha }_p=\frac{2(mg-2ma-f)}{mR}$ $---\left(3\right)$

Now , Acceleration of pulley $\left(a\right)=R{\alpha }_p$

$⟹$ ${\alpha }_p=\frac{a}{R}$

$\frac{2(mg-2ma-f)}{mR}=aR$ ( from 3 )

$\therefore$ $ma=\frac{2(mg-f)}{5}$ $----\left(4\right)$

Now , On equalling the Torque for cylinder we get ,

$2Rf=I_c{\alpha }_c$

$⟹$ ${\alpha }_c=\frac{f}{mR}$ $--------\left(5\right)$ $(\because I_c=2m{R}^2)$

Now , Acceleration of cylinder $\left(a\right)$ = $2R{\alpha }_c$

$⟹$ ${\alpha }_c=\frac{a}{2R}$

$\frac{f}{mR}=\frac{a}{2R}$ ( from 5 )

$⟹$ $ma=2f$

$\frac{2(mg-f)}{5}=2f$ ( from 4 )

$\therefore$ $\underset{–}{f=\frac{mg}{6}}$

Hence , $\underset{–}{OptionD\left(\frac{mg}{6}\right)iscorrect.}$