Simple Harmonic (3).

Classical Mechanics Level 2

A uniform rod of length ℓ \ell sits on top of a hemisphere of radius R R as shown in the figure. The friction between hemisphere is sufficient enough to prevent slipping. At this instant, the rod is placed in equilibrium. The rod is now given a slight angular displacement towards either side and it executes SHM. If the time period of this SHM is given by

T = π ℓ a R g T = \dfrac{\pi \ell}{\sqrt{aRg}}

find the value of a a .

Details: g = 9.8 ms − 2 g = 9.8 \text{ms}^{-2} denotes the acceleration due to gravity.


The answer is 3.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Suhas Sheikh
Jun 8, 2018

Treat as a physical pendulum about the contact point with the hemisphere and just apply formulae Sufficient for the problem

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