Question

The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is

 $$\begin{gathered} \mathbf{A}.\mathbf{kx} \\ \mathbf{B}.\mathbf{2}\mathbf{kx} \\ \mathbf{C}.{\displaystyle \frac{\mathbf{2}\mathbf{kx}}{\mathbf{3}}} \\ \mathbf{D}.{\displaystyle \frac{\mathbf{4}\mathbf{kx}}{\mathbf{3}}} \end{gathered}$$

Answer 1

Hint: Net Force acting on a body is given by the formula:
$${\overrightarrow{F}}_{Net}=m.\stackrel{\rightharpoonup }{a}$$
Where,
$${\overrightarrow{F}}_{Net}$$ is the net force acting on the body
m is the mass of the body
$$\stackrel{\rightharpoonup }{a}$$ is the acceleration of the body
Moment of Force is called Torque.
Net Torque acting on a body is given by the formula,
$${\overrightarrow{\tau }}_{Net}=I\overrightarrow{\alpha }$$
Where,
$${\overrightarrow{\tau }}_{Net}$$ is the net Torque acting on the body
I is the moment of inertia of the body
$$\overrightarrow{\alpha }$$is the angular acceleration of the body
Also,
$${\overrightarrow{\tau }}_{Net}=\overrightarrow{F}\times R$$
Moment of inertia of a disk is given by the formula,
$I={\displaystyle \frac{1}{2}}M{R}^2$

Using all the above formulas, we can easily compute the result.

Complete step by step solution: Net Force acting on a body is given by the formula:
$${\overrightarrow{F}}_{Net}=m.\stackrel{\rightharpoonup }{a}$$
Where,
$${\overrightarrow{F}}_{Net}$$ is the net force acting on the body
m is the mass of the body
$$\stackrel{\rightharpoonup }{a}$$ is the acceleration of the body
We will insert $$(-2kx+F)$$in the place of $${\overrightarrow{F}}_{Net}$$
$$-2kx+F=M{a}_c$$ Equation 1
Where,
k is the spring constant
x is the distance by which the spring has been stretched
F is the friction force on the disk
M is the mass of disk
$$a_c$$ is the acceleration of centre of mass of the disk
Net Torque acting on a body is given by the formula,
$${\overrightarrow{\tau }}_{Net}=I\overrightarrow{\alpha }$$ Equation 2
Where,
$${\overrightarrow{\tau }}_{Net}$$ is the net Torque acting on the body
I is the moment of inertia of the body
$$\overrightarrow{\alpha }$$is the angular acceleration of the body
Additionally, Torque is also calculated as follows,
$${\overrightarrow{\tau }}_{Net}=\overrightarrow{F}\times R$$ Equation 3
Where,
R is the distance of the Force from the centre of mass of the body

Now combining equations 2 and 3,
We get,
$\overrightarrow{F}\times R=I\overrightarrow{\alpha }$
$\overrightarrow{F}={\displaystyle \frac{I\overrightarrow{\alpha }}{R}}$ Equation 4
In Pure Rolling condition,
$$a_c={\overrightarrow{\alpha }}_{c}R$$
Moment of inertia of a disk is given by the formula,
$I={\displaystyle \frac{1}{2}}M{R}^2$
Inserting the values of $\alpha$ and I in equation 4,
We get,
$$F={\displaystyle \frac{{\displaystyle \frac{1}{2}}M{R}^2}{R}}\times {\displaystyle \frac{a_c}{R}}$$
$=>F={\displaystyle \frac{1}{2}}M{a}_c$
Inserting the value of F in equation 1,
We get,
$=>-2kx+{\displaystyle \frac{1}{2}}M{a}_c=-M{a}_c$
$=>{\displaystyle \frac{1}{2}}M{a}_c+M{a}_c=2kx$
$=>{\displaystyle \frac{3}{2}}M{a}_c=2kx$
$=>M{a}_c={\displaystyle \frac{4}{3}}kx$
$=>M{a}_c=-{\displaystyle \frac{4}{3}}kx$

Hence, Option (D) is correct.

Note:
We have used a negative sign because the disk has been displaced away from the equilibrium position. Hence this force will tend to bring the disk back to its initial position.