Question:

On imparting an initial velocity \(v_0\), a ball begins to move in horizontal circle of radius \(R\) on horizontal plane. If the coefficient of friction between the ball and plane is \(\mu\), then the time required by ball to come to rest is: center
center

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Whenever friction alone slows a body on a horizontal surface: \[ a=\mu g \] This is one of the most important standard results in mechanics.
Updated On: Jun 17, 2026
  • \(\dfrac{v_0^2}{\mu g}\)
  • \(\dfrac{\mu g}{v_0}\)
  • \(\dfrac{v_0\mu}{g}\)
  • \(\dfrac{v_0}{\mu g}\)
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The Correct Option is D

Solution and Explanation

Concept: Friction opposes motion and produces retardation. Frictional force: \[ f=\mu N \] On a horizontal plane: \[ N=mg \] Hence, \[ f=\mu mg \] Retardation: \[ a=\frac{f}{m} \] \[ =\mu g \] Since friction is constant, motion is uniformly retarded. Using first equation of motion: \[ v=u-at \] At stopping point: \[ v=0 \]

Step 1: Apply equation of motion. \[ 0=v_0-\mu gt \] \[ \mu gt=v_0 \] \[ t=\frac{v_0}{\mu g} \] Therefore, \[ \boxed{\frac{v_0}{\mu g}} \]
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