Question:
A small spherical ball of radius ' r ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its time period of oscillation is proportional to ( g = acceleration due to gravity)}
A small spherical ball of radius ' r ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its time period of oscillation is proportional to ( g = acceleration due to gravity)}
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The path of the center of mass determines the effective length of oscillation.
Updated On: Apr 26, 2026
- $\sqrt{\frac{R}{g}}$
- $\sqrt{\frac{r}{g}}$
- $\sqrt{\frac{R-r}{g}}$
- $\sqrt{\frac{R+r}{g}}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The ball moves along the arc of a circle of radius $R$. The center of the ball moves in an arc of radius $(R-r)$.
Step 2: Effective Length
The motion is analogous to a simple pendulum where the effective length $L = R - r$.
Step 3: Time Period Formula
$T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{R-r}{g}}$.
Thus, $T \propto \sqrt{\frac{R-r}{g}}$.
Final Answer: (C)
The ball moves along the arc of a circle of radius $R$. The center of the ball moves in an arc of radius $(R-r)$.
Step 2: Effective Length
The motion is analogous to a simple pendulum where the effective length $L = R - r$.
Step 3: Time Period Formula
$T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{R-r}{g}}$.
Thus, $T \propto \sqrt{\frac{R-r}{g}}$.
Final Answer: (C)
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