GATE 2022 ME (ME1) – Question 32
A rigid uniform annular disc is pivoted on a knife edge A in a uniform gravitational field as shown, such that it can execute small amplitude simple harmonic motion in the plane of the figure without slip at the pivot point. The inner radius $r$ and outer radius $R$ are such that $r^2 = R^2/2$, and the acceleration due to gravity is $g$. If the time period of small amplitude simple harmonic motion is given by $T = \beta\pi\sqrt{R/g}$, where $\pi$ is the ratio of circumference to diameter of a circle, then $\beta =$ ________ (round off to 2 decimal places).
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Correct answer: 2.65 to 2.67
Explanation
The knife edge A is on the inner edge, a distance $r$ from the centre G. The moment of inertia about G is $I_G = \frac{1}{2}m(R^2 + r^2) = 0.75mR^2$, and about A it is $I_A = I_G + mr^2 = 0.75mR^2 + 0.5mR^2 = 1.25mR^2$. For small oscillations $\omega^2 = \frac{mgr}{I_A} = \frac{g(R/\sqrt{2})}{1.25R^2} = 0.5657\frac{g}{R}$. The period is $T = \frac{2\pi}{\omega} = \frac{2}{\sqrt{0.5657}}\pi\sqrt{\frac{R}{g}} = 2.66\pi\sqrt{R/g}$.