Question:medium

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: May 7, 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: Understanding the Concept:
For a small object oscillating in a spherical bowl or on a curved surface, the system acts like a simple pendulum. The effective length of this "pendulum" is the distance from the center of curvature of the surface to the center of mass of the oscillating object.
Step 2: Key Formula or Approach:
The time period \(T\) of a simple pendulum or equivalent system is:
\[ T = 2\pi\sqrt{\frac{L_{eff}}{g}} \]
Where \(L_{eff}\) is the distance from the point of suspension (center of curvature) to the center of the ball.
Step 3: Detailed Explanation:
The radius of curvature of the surface is \(R\).
The radius of the spherical ball is \(r\).
When the ball is at the bottom of the surface, its center is at a distance \(r\) from the surface.
The center of curvature of the surface is at a distance \(R\) from the surface.
The distance between the center of the surface and the center of the ball is:
\[ L_{eff} = R - r \]
Therefore, the time period of oscillation is:
\[ T = 2\pi\sqrt{\frac{R-r}{g}} \]
The proportionality is:
\[ T \propto \sqrt{\frac{R-r}{g}} \]
Step 4: Final Answer:
Its time period of oscillation is proportional to \(\sqrt{\frac{R-r}{g}}\).
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