MHT CET · Physics · Rotational Motion

Three point masses each of mass ‘m’ are kept at the corners of an equilateral triangle of side ‘L’. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to $(c o s 30^{o} = s i n 60^{o} = \frac{\sqrt{3}}{2})$

A $\sqrt{L}$
B $L^{3 / 2}$
C $L$
D $L^{2}$

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Answer & Solution

Correct Answer

(D) $L^{2}$

Step-by-step Solution

Detailed explanation

The situation can be shown below,

Where, O is the centroid of equivalent triangle.
The length AO can be found by

A O = \sqrt{O D^{2} + A D^{2}}

= \sqrt{{(\frac{1}{3} B D)}^{2} + A D^{2}} = \sqrt{\frac{1}{9} \times {(\frac{\sqrt{3}}{2} L)}^{2} + \frac{L^{2}}{4}}

= \sqrt{\frac{1}{12} L^{2} + \frac{L^{2}}{4}} = \frac{L}{\sqrt{3}}

The moment of inertia about O is given by

I = M R^{2}

= m \times {(\frac{L}{\sqrt{3}})}^{2} = \frac{m L^{2}}{3}

According to law of conservation of angular momentum,

I_{ω} =

Constant

I \times \frac{2 π}{T}

= constant

\Rightarrow T \propto L^{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Three point masses each of mass ‘m’ are kept at the corners of an equilateral triangle of side ‘L’. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to cos30o=sin60o=32

Step-by-step Solution