Question

A solid sphere of radius R is set into motion on a rough horizontal surface with a linear speed \[{v_o}\] in forward direction and an angular velocity \[{\omega _o} = {v_o}/2R\] in counter coefficient of friction is\[\mu \], then find the time after which sphere starts pure rolling.


A. \[{t_o} = \dfrac{{3{v_o}}}{{7\mu g}} \\ \]
B. \[{t_o} = \dfrac{{2{v_o}}}{{7\mu g}} \\ \]
C. \[{t_o} = \dfrac{{4{v_o}}}{{7\mu g}} \\ \]
D. \[{t_o} = \dfrac{{{v_o}}}{{7\mu g}}\]

Answer 1

Hint:Frictional force is a special type of force which is usually against the direction of motion. In the above problem, the frictional force is between the rough horizontal surface and a solid sphere. For pure rolling, the rotational and translational velocity should be equivalent. The time taken for that velocity will give the time for rolling.

Formula Used:
Frictional force,
\[f = \mu Ma\]
where \[\mu \]= coefficient of friction, M = mass of the sphere and a = acceleration. This acceleration can be g (acceleration due to gravity) if the motion is vertical or about an inclined plane.
\[v = \omega R\]
where v = linear velocity, \[\omega \]= angular velocity and R is the radius of the sphere.

Complete step by step solution:
Given: A solid sphere of mass m and radius R with initial linear velocity \[{v_o}\] and initial angular velocity, \[{\omega _o} = {v_o}/2R\]. The coefficient of friction is given as \[\mu \].Let the frictional force applied be f and time for pure rolling be \[{t_o}\]. Since friction acts against the direction of motion, it will cause deacceleration. Let it be $a$.
Then,
\[ma = f = \mu mg\]
Then,
\[a = \mu g\]---- (1)

According to the question, applying equations of motion,
Linear velocity at time \[{t_o}\]is
\[v = {v_o} - at = {v_o} - \mu g{t_o}\]---- (2)
Moment of inertia for a solid sphere for rotation about its axis, \[I = \dfrac{2}{5}m{R^2}\].
We know that,
\[I\alpha = fR\]where \[\alpha \]is the angular acceleration
So, \[\alpha = \dfrac{{fR}}{I} = \dfrac{{fR}}{{\left( {\dfrac{2}{5}m{R^2}} \right)}}\]
\[\alpha = \dfrac{{5\mu g}}{{2R}}\]---- (3)
At a given time \[{t_o}\], the angular velocity will be
\[\omega = - {\omega _0} + \alpha t = - \dfrac{{{v_o}}}{{2R}} + \dfrac{{5\mu g}}{{2R}}{t_o}\]---(4)

Negative sign of the angular velocity indicates spin in the reverse direction while positive sign on acceleration is because it is in forward direction. For pure rolling,
\[v = \omega R\]--- (5)
Thus,
\[{v_o} - \mu gt = ( - \dfrac{{{v_o}}}{{2R}} + \dfrac{{5\mu g}}{{2R}}{t_o})R\]
\[\Rightarrow 3{v_o} = 7\mu g{t_o}\]
\[\therefore {t_o} = \dfrac{{3{v_o}}}{{7\mu g}}\]
Thus, the time after which pure rolling starts is \[{t_o} = \dfrac{{3{v_o}}}{{7\mu g}}\].

Hence option A is the correct answer.

Note: For pure rolling, the rotational acceleration and the acceleration of the centre of mass should have an equivalence of the form \[R\alpha = {a_{CM}}\]. The angular velocity in case of deacceleration will have an opposing direction to the direction of acceleration in order to bring the motion of the body to stop. Hence it will have a negative sign.