A small rubber ball of radius $\mathit{r}$hits a rough horizontal floor such that its speed$\mathit{v}$just before striking the floor at location A makes an angle of $\mathbf{60}\mathbf{°}$with the horizontal and also has back spin with angular speed $\mathit{\omega }$. It is observed that the ball repeatedly bounces from A to B, then from B back to A, etc. Assuming perfectly elastic impact determine (a) the required magnitude of $\mathit{\omega }$of the back spin in terms of $\mathit{v}$and $\mathit{r}$, and (b) the minimum magnitude of co-efficient of static friction${\mathit{\mu }}_{\mathbf{s}}$ to enable this motion. Hint: Notice that the direction of $\overrightarrow{\mathbf{\omega }}$ flips in each collision.

It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object. It is a vector quantity, which implies that here along with magnitude, the direction is also considered.

If a ball is falling with a speed and hits the surface, it will rebound to some vertical height and again fall back. In this type of motion, the horizontal component of velocity remains the same.

The following figure represents motion diagram of the ball

From the figure, the velocity vector of the ball makes an angle of$60°$ with the horizontal. When the ball hits the surface, it rebounds by making an angle of$30°$ with the horizontal because as the impact is perfectly elastic. Here, $v$and $v\text{'}$represent the velocities of the ball before and after the collision. $f_s$and $F_C$represents the static frictional force and the centripetal force due to spin of the ball and which are acting opposite to each other.

Equate the horizontal components of the velocities of the ball before and after the collision.

$\begin{array}{rcl}\cos 60°& =& v\text{'}\cos 30°\\ v\text{'}& =& \frac{v\left(\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\\ & =& \frac{v}{\sqrt{3}}\\ & & \end{array}$

Determine the angular velocity of the back spin of the ball by using the following formula:

$v\text{'}=r\omega$

Here, $r$is the radius of the ball and $\omega$is the angular velocity of the spinning ball.

Substitute $\frac{v}{\sqrt{3}}$for $v\text{'}$

$\begin{array}{rcl}\frac{v}{\sqrt{3}}& =& r\omega \\ \omega & =& \frac{v}{\left(\sqrt{3}\right)r}\\ & & \end{array}$

Therefore, the required value of$\omega$ of the back spin is $\omega =\frac{v}{\sqrt{3}r}$.

The spinning force balances the static friction to maintain equilibrium as the impact is perfect elastic.

$f_s=F_C$

The expression for $f_s$is given as follows:

$f_s={\mu }_{s}mg$

Here, ${\mu }_s$is coefficient of static friction, $m$is the mass of the ball and$g$is acceleration due to gravity.

The expression for $F_C$is given as follow:

$F_C=mr{\omega }^2$

Substitute${\mu }_{s}mg$ for $f_s$ and $mr{\omega }^2$for $F_C$in the equation $f_s=F_C$.

${\mu }_{s}mg=mr{\omega }^2$

Rearrange the equation for ${\mu }_s$and substitute $\frac{v}{\sqrt{3}r}$for $\omega$.

$\begin{array}{rcl}{\mu }_s& =& \frac{r{\left(\frac{v}{\sqrt{3}r}\right)}^2}{g}\\ & =& \frac{v^2}{3gr}\\ & & \end{array}$

Hence, the minimum magnitude of coefficient of static friction to enable this motion is $\frac{v^2}{3gr}$.