Question

A uniform wooden rod of length L and specific gravity S is free to rotate in the vertical plane about the end A. The rod is released from the horizontal position inside a liquid of specific gravity 1.5 S as shown in the figure. The instantaneous acceleration at the end B is:

a. \[g\]
b. \[\dfrac{g}{4}\]
c. \[\dfrac{3g}{4}\]
d. \[\dfrac{3g}{8}\]

Answer 1

Hint: When the wooden rod is inserted in the liquid with the side A fixed, the point B will rise upwards due to the buoyant force \[\left( {{F}_{b}} \right)\]. Now we will equate the product of the force with the distance with the moment of inertia product with angular acceleration or also known as torque. The force in the above question can be converted in terms of specific gravity (S), length and area.

The formula for the torque is given as:
\[\text{Torque}=I\alpha \]
The formula for the force can be written as:
\[F=S.A.L.g\]
where I is the moment of inertia and \[\alpha \] is the angular acceleration, S is the specific gravity, A is the area of the rod, L is the length of the rod and g is the gravity as \[10\text{ }m{{s}^{-2}}\].

Complete step by step answer:
When inserting the rod in the water about point A, the free point of B rises upward due to the buoyant force acting on the rod due to the higher pressure form in the water pushing the rod on the surface of the water. The formula for the torque is given as:
\[\text{Torque}=I\alpha \]

The torque acted due to the buoyant force in the middle of the rod is given as the product of force by the length of the rod by half length.
Torque acted in the middle of the rod is \[\left( {{F}_{b}}-mg \right)\dfrac{L}{2}\]

Converting the force in terms of specific gravity, area and length as:
\[\Rightarrow \left( {{F}_{b}}-mg \right)\dfrac{L}{2}\]
\[\Rightarrow \left( S\times A\times L\times g-1.5\times S\times A\times L\times g \right)\dfrac{L}{2}\]

Equating both the torques i.e. the torque experienced in the middle of the rod by the torque experience on the outer due to buoyant force acting on the rod is:
\[\Rightarrow \left( S\times A\times L\times g-1.5\times S\times A\times L\times g \right)\dfrac{L}{2}\]
Equating the torques on both RHS and LHS, we get the value of the angular acceleration as:
\[\Rightarrow I\alpha =\left( S\times A\times L\times g-1.5\times S\times A\times L\times g \right)\dfrac{L}{2}\] ( The value of the moment of inertia of a rod about its end is \[I=\dfrac{mL}{3}\])
\[\Rightarrow \dfrac{mL}{3}\alpha =\left( S\times A\times L\times g-1.5\times S\times A\times L\times g \right)\dfrac{L}{2}\]
\[\Rightarrow \alpha =mg\dfrac{3L}{2mL}-mg\dfrac{3\times 1.5\times L}{2mL}\]
\[\Rightarrow \alpha =g\dfrac{3}{2}-g\dfrac{3\times 1.5}{2}\]
\[\Rightarrow \alpha =\dfrac{3}{4}g\]
Therefore, The angular acceleration or acceleration on the point B is \[\alpha =\dfrac{3g}{4}\].

Hence, the correct answer is option (B).

Note: The student may go wrong while converting the torque from product of mass and gravity to specific gravity, area, and length. The term specific gravity and gravity are two different terms and not the same and the torque is the product of force and displacement of the object (angular). Therefore, there will be two displacement terms i.e. the torque in terms of specific gravity is \[\left( S\times A\times L\times g \right)\times \dfrac{L}{2}\].