Question

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.99 J of work to set the ball spinning from rest, at what angular speed ω will the ball rotate? Model a basketball as a thin-walled hollow sphere. For a men's basketball, the ball has a circumference of 0.749 m and a mass of 0.624 kg .

Answer 1

To solve this problem it is necessary to apply the concepts related to rotational kinetic energy, the definition of the moment of inertia for a sphere and the obtaining of the radius through the circumference. Mathematically kinetic energy can be given as:

$KE= I\omega^2$

Where,

I = Moment of inertia

$\omega =$ Angular velocity

According to the information given we have that the radius is

$\Phi= 2\pi r$

$0.749m = 2\pi r$

$r = 0.1192m$

With the radius obtained we can calculate the moment of inertia which is

$I = \frac{2}{3}mr^2$

$I = \frac{2}{3}(0.624)(0.1192)^2$

$I = 5.91*10^{-3} kg \cdot m^2$

Finally, from the energy equation and rearranging the expression to obtain the angular velocity we have to

$\omega = \sqrt{\frac{2KE}{I}}$

$\omega = \sqrt{\frac{2(1.99)}{5.91*10^{-3}}}$

$\omega = 25.95rad/s$

Therefore the angular speed will the ball rotate is 25.95rad/s