Question

The net reaction of the disc on the block is:

\left( A \right)m{\omega ^2}R\sin \omega t\hat j - mg\hat k \\\ \left( B \right)\dfrac{1}{2}m{\omega ^2}R\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)\hat j + mg\hat k \\\ \left( C \right)\dfrac{1}{2}m{\omega ^2}R\left( {{e^{2\omega t}} - {e^{ - 2\omega t}}} \right)\hat j + mg\hat k \\\ \left( D \right) - m{\omega ^2}R\cos \omega t\hat j - mg\hat k \\\

Answer 1

Hint : In order to solve this question, we are going to first analyze the figure that is given in the question, after that the expression for the rotational force is written, in which the value of the rotational force is put and simplified for the internal force on the disc, after which the reaction force is calculated.
The rotational force on the disc is given by
${F_{rot}} = {F_{in}} + 2m\left( {{v_{rot}}\hat i} \right) \times \omega \hat k + m\left( {\omega \hat k \times r\hat i} \right) \times \omega \hat k$
The rotational force is,
${F_{rot}} = m{\omega ^2}r\hat i$ .

Complete Step By Step Answer:
As we can see in the figure, we are given with a disc of radius $R$ , which is rotating with the angular velocity equal to $\omega$ ,
The block of mass, $m$ , is given at a distance $\dfrac{R}{2}$ from the centre of the radius.
The rotational force of the disc depends on the internal force present on the disc. It is given by the equation,
${F_{rot}} = {F_{in}} + 2m\left( {{v_{rot}}\hat i} \right) \times \omega \hat k + m\left( {\omega \hat k \times r\hat i} \right) \times \omega \hat k$
We know that the general expression for the rotational force is,
${F_{rot}} = m{\omega ^2}r\hat i$
Putting this value in the above equation, we get
$m{\omega ^2}r\hat i = {F_{in}} + 2m{v_{rot}}\omega \left( { - \hat j} \right) + m{\omega ^2}r\hat i$
Hence, simplifying this equation, we get the value of internal force equal to
${F_{in}} = 2m{v_{rot}}\omega \hat j$
The radius $r$ can be written as
$r = \dfrac{R}{4}\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)$
The velocity can be found by differentiating the above equation,
${v_r} = \dfrac{{dr}}{{dt}} = \dfrac{{R\omega }}{4}\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)$
Putting this value of velocity in the expression for the internal force, we get
{F_{in}} = \dfrac{{2m{\omega ^2}R}}{4}\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)\hat j \\\ \Rightarrow {F_{in}} = \dfrac{{m{\omega ^2}R}}{2}\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)\hat j \\\
Thus, the reaction force becomes,
${F_{reaction}} = \dfrac{{m{\omega ^2}R}}{2}\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)\hat j + mg\hat k$
Hence, the option $\left( B \right)\dfrac{1}{2}m{\omega ^2}R\left( {{e^{\omega t}} - {e^{ - \omega t}}} \right)\hat j + mg\hat k$ is the correct answer.

Note :
The force of rotation on a disc depends on the internal force present inside the disc, the velocity of the rotation of the disc and the moment of inertia of a disc. These three quantities form an expression which is further simplified in order to get the internal force in the disc which along with gravity gives reaction force.