Question

A cotton reel placed on a rough surface has an inner radius $r$ and outer radius $R$ . Mass of the reel is $M$ and moment of inertia about longitudinal rotational axis is $I.$ A force $F$ is applied at the free end of thread wrapped on the reel as shown in figure. If the reel moves without sliding, then

• A. $a_{reel}=\frac{FR(R-r)}{I+M{R}^2}$
• B. $a_{reel}=\frac{FR(R+r)}{I+M{R}^2}$
• C. ${\alpha }_{reel}=\frac{F(R+r)}{I+M{R}^2}$
• D. $F_{friction}=\frac{F(1+MRr)}{I+M{R}^2}$

Answer 1

The correct option is D $F_{friction}=\frac{F(1+MRr)}{I+M{R}^2}$


As the body rolls, point $P$ will be instantaneous center of rotation (ICR). There is clockwise torque about ICR. Hence, center of mass will have acceleration in right direction and the reel will rotate in clockwise direction.

The equations of motion are
$\sum F_x=F-f=Ma$ .....(i)
$\sum F_y=N-Mg=0$ ........(ii)
$\sum \tau =f\times R-F\times r=I\alpha$ .....(iii)

As the reel rolls without slipping, $a=R\alpha$

From eqs. (i) and (iii),
$\frac{F-f}{M}=\frac{R(f\times R-F\times r)}{I}$
$f=\frac{F(1+MRr)}{I+M{R}^2}$

Force of friction comes out to be positive, therefore our assumption about the direction of friction force was correct.

On substituting the expression for $f$ in (i)
we obtain $a=\frac{F}{M}[1-\frac{(1+MRr)}{I+M{R}^2}]$

Also, $\alpha =\frac{a}{R}=\frac{F}{MR}[1-\frac{(1+MRr)}{I+M{R}^2}]$