Question

Is it possible to choose r, so that a is greater than $\dfrac{F}{M}$? How?

Answer 1

Hint: Think how we can make the acceleration of the disc greater than $\dfrac{F}{M}$. If the net force is greater than F the acceleration will increase. When a force is exerted on a rigid body, other than translation motion, there is rotational motion too.
Formulas used:
F=ma
$\tau =Fr$
$\tau =I\alpha$
${{a}_{T}}=R\alpha$

Complete step by step answer:
‘a’ is the acceleration of the centre of mass of the disc i.e. the centre of the disc. If we use newton’s second law F=ma, $a=\dfrac{F}{M}$.
Therefore, you may think that a cannot be greater than $\dfrac{F}{M}$ . However, what if another force acts on the disc, in the direction along the force F. Then the force will increase and therefore the acceleration will become greater than $\dfrac{F}{M}$.
When the force F is applied on the disc, due to relative motion between the disc and the ground, at the point of contact, friction force (f) will come into play and this frictional force should be in the same direction of force F, as explained above.

Therefore, the net force will be $F+f=ma$. ………(1)
These two forces will create a torque $\tau =Fr-fR$.
Torque is also equal to $I\alpha$, where I is the moment of inertia of the disc about its centre and $\alpha$ is the angular acceleration of the disc. Moment of inertia for a disc about its centre is equal to $I=\dfrac{1}{2}M{{R}^{2}}$.
Therefore, $\tau =Fr-fR=I\alpha$
$\Rightarrow Fr-fR=\dfrac{1}{2}M{{R}^{2}}\alpha$ …….(2)
We can rewrite equation (2) as
$Fr-fR=\dfrac{1}{2}MR\left( R\alpha \right)$ ……..(3)
Due to angular acceleration there is tangential acceleration ${{a}_{T}}=R\alpha$.
Due to friction, the relative at point of contact will be zero. Therefore, ${{a}_{T}}=a\Rightarrow R\alpha =a$ and from equation (1), $a=\dfrac{F+f}{M}$.
$\Rightarrow R\alpha =\dfrac{F+f}{M}$
Substitute the value of $R\alpha$ in equation (3).
Therefore, $Fr-fR=\dfrac{1}{2}MR\left( \dfrac{F+f}{M} \right)$
$\Rightarrow 2Fr-2fR=RF+Rf$
$\Rightarrow F(2r-R)=3Rf$
Therefore, $\Rightarrow f=\dfrac{F(2r-R)}{3R}$. The required condition is $F+f >F$. This implies $f >0$.
Therefore, $\dfrac{F(2r-R)}{3R} >0$
F and 3R are positive, then (2r-R) must be positive for the expression to be positive.
$\Rightarrow 2r-R > 0$
$\Rightarrow r >\dfrac{R}{2}$.
Therefore, for a to be greater than $\dfrac{F}{M}$, r must be greater than half of the radius (R) of the disc.

Note: Let us see what will happen if $r < \dfrac{R}{2}$. If r is less than the radius of the disc, f will be negative. That means the frictional force will be in the opposite direction of force F. Therefore the net force will be $F-f$. This means the net force will be less than F. Therefore the acceleration of the disc will be less than $\dfrac{F}{M}$.