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A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle$\theta $ with the vertical is
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$\eqalign{
  & {\text{A}}{\text{. }}\dfrac{{3g}}{{2l}}\cos \theta \cr
  & {\text{B}}{\text{. }}\dfrac{{2g}}{{3l}}\cos \theta \cr
  & {\text{C}}{\text{. }}\dfrac{{3g}}{{2l}}\sin \theta \cr
  & {\text{D}}{\text{. }}\dfrac{{2g}}{{3l}}\sin \theta \cr} $

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Last updated date: 27th Mar 2024
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MVSAT 2024
Answer
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Hint:When the rod is released, it will perform rotational motion due to torque. Find the value of torque with relation to angular acceleration. Also deduce the expression for the moment of inertia about the pivot. Equate all the formed mathematical expressions to get the required answer.

Formula Used:
Torque, $\tau = I\alpha = r \times F$
Parallel axis theorem:
$\eqalign{
  & I = {I_{cm}} + M{d^2} \cr
  & {\text{where I is the moment of inertia about the parallel axis,}} \cr
  & {I_{cm}}{\text{ is the moment inertia about the center,}} \cr
  & {\text{M is the mass of the rod,}} \cr
  & {\text{d is the distance between the two axes}}{\text{.}} \cr} $



Complete step by step solution:
Given is a system of a slender uniform rod having mass M and length l. It can rotate in the vertical plane; as it is pivoted at one end only. Additionally, there is negligible friction at the pivot.
When the free end of the rod is released at it makes an angle$\theta $ with the vertical, then its weight acts vertically downwards. This weight is the only factor contributing for the rotational torque because the other two forces acting at the pivot have zero distance from the point of application of force, thus their torque will be zero as torque is a cross product of force and the perpendicular distance of the line of action of force.
Thus torque acting on the rod will be:
$\tau = I\alpha \cdots \cdots \cdots \cdots \left( 1 \right)$
Also, $\tau = r \times F \cdots \cdots \cdots \cdots \left( 2 \right)$
Now, as the rod is uniform in nature, so its weight must act directly at its midpoint which will be situated at half of its length l.
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So, from equation (2) we have:
$\eqalign{
  & \tau = r \times F \cr
  & \tau = Mg\dfrac{l}{2}\sin \theta \cdots \cdots \cdots \cdots \left( 3 \right) \cr} $
Both equation (1) and equation (2) represent the same physical quantities, so equating both, we get:
$\eqalign{
  & r \times F = I\alpha \cr
  & \Rightarrow Mg\dfrac{l}{2}\sin \theta = I\alpha \cdots \cdots \cdots \left( 4 \right) \cr} $
The moment of inertia of the rod about the pivot point will be given as:
$\eqalign{
  & I = {I_{cm}} + M{d^2} \cr
  & I = \dfrac{{M{l^2}}}{{12}} + M{\left( {\dfrac{l}{2}} \right)^2}{\text{ }}\left[ {\because {I_{cm}} = \dfrac{{M{l^2}}}{{12}}{\text{ }}} \right] \cr
  & I = \dfrac{{M{l^2}}}{{12}} + \dfrac{{M{l^2}}}{4} \cr
  & I = \dfrac{{M{l^2} + 3M{l^2}}}{{12}} \cr
  & I = \dfrac{{M{l^2}}}{3} \cr} $
Substituting the value of moment of inertia in equation (4), we get:
$\eqalign{
  & Mg\dfrac{l}{2}\sin \theta = \dfrac{{M{l^2}}}{3}\alpha \cr
  & \Rightarrow \alpha = \dfrac{{3gl\sin \theta }}{{2{l^2}}} \cr
  & \therefore \alpha = \dfrac{{3g}}{{2l}}\sin \theta \cr} $

Therefore, the correct option is C i.e., the angular acceleration of the rod when it makes an angle$\theta $ with the vertical is $\dfrac{{3g}}{{2l}}\sin \theta $


Note:Students can make errors while applying the standard equation of moment of inertia about the center and perpendicular to the axis. This in turn would result in the wrong expression for the moment of inertia about the pivot by parallel axis theorem. Avoid such mistakes. Additionally, if the given rod is non-uniform take a small element dx with mass dm, and integrate the same to get the mass of the entire rod.

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