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
Show Hint
Whenever friction alone slows a body on a horizontal surface:
\[
a=\mu g
\]
This is one of the most important standard results in mechanics.
Step 1: Find what slows the ball. The ball moves on a flat surface, so the only thing that slows it down is friction. The size of the friction force decides how fast the speed drops.
Step 2: Write the friction force. On a flat plane the normal force equals the weight. \[ f = \mu N = \mu m g \] Step 3: Get the slowing rate. Acceleration is force over mass, and it points backward. \[ a = \frac{f}{m} = \mu g \] The mass cancels, which is a neat result. Step 4: Note that this slowing is steady. Since $\mu g$ is constant, the speed falls at a steady rate, so we can use the simple straight line motion formula. \[ v = v_0 - a t \] Step 5: Apply the stopping condition. The ball stops when $v = 0$. \[ 0 = v_0 - \mu g\, t \] Step 6: Solve for the time. \[ t = \frac{v_0}{\mu g} \] \[ \boxed{\dfrac{v_0}{\mu g}} \]
