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At a bicycle repair shop, a bicycle tire of mass M and radius R is suspended from a peg on the wall. The moment of inertia of the tire around the peg is 2MR2 . If the tire is displaced from equilibrium and starts swinging back and forth, what will be its frequency of oscillation

Respuesta :

The frequency of oscillation given the moment of inertia of the tire around the peg will be f = 1/2π✓(g/2R).

How to calculate the frequency of oscillation?

Given I = 2MR²

Torque = Mgsin(theta)R

Here, w = 2Ï€f.

f = 1/2π✓(MgR/I)

f = 1/2π✓(MgR/2MR²)

f = 1/2π(✓g/2R)

Therefore, the frequency of oscillation is f = 1/2π(✓g/2R).

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