Oscillations

JEE Main level, single correct, medium

147 of 194

Question 147 of 194

A uniform disc of radius RR and mass MM is free to oscillate about the axis AA as shown in the figure. For small oscillations the time period is ______.
(
gg is acceleration due to gravity)

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Answer: (A) 2π5R4g2\pi\sqrt{\dfrac{5R}{4g}}

From the given figure, the axis of rotation is a horizontal line passing through the point AA on the edge of the disc. The circular arrow around the axis indicates that the disc oscillates out of its plane, meaning the axis of rotation is a tangent to the disc in its own plane.

The moment of inertia of a uniform disc about its diameter is
Icm=14MR2I_{cm} = \dfrac{1}{4}MR^2.

Using the parallel axis theorem, the moment of inertia of the disc about the given axis of rotation is:

I=Icm+MR2=14MR2+MR2=54MR2I = I_{cm} + MR^2 = \dfrac{1}{4}MR^2 + MR^2 = \dfrac{5}{4}MR^2

The distance from the center of mass of the disc to the axis of rotation is
d=Rd = R.

The time period of a physical pendulum is given by:

T=2πIMgdT = 2\pi\sqrt{\dfrac{I}{Mgd}}

Substituting the values of
II and dd:
T=2π54MR2MgR=2π5R4gT = 2\pi\sqrt{\dfrac{\dfrac{5}{4}MR^2}{MgR}} = 2\pi\sqrt{\dfrac{5R}{4g}}

Answer:
2π5R4g2\pi\sqrt{\dfrac{5R}{4g}}

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