Question

When a pitcher throws a baseball or softball, the rotation speed is important when it comes to making the ball "break," or curve away from its expected motion. A softball that has a mass of 0.200 kg and a radius of 4.77 cm can rotate at a rate of 1.4410pm on its way to home plate, causing the ball to curve! Suppose this softball is rotating around an axis through its center at this extreme rate. What is the rotational kinetic energy of the softball as it travels toward home plate?

Answer 1

Solution :-

Given, Mass (M) = 0.200 kg

Radius = 4.77cm

Moment of inertia of ball = $\dpi{80} \frac{2}{5} mR^2$

=(2×(0.200)×(4.77×10^-2)^2)/5

=1.82x10^-4 kg-m_²

$\dpi{100} \omega$ = 1440 rpm = 1440×2×3.14/60 rad/sec = 150.72 rad/sec

Rotational Kinetic energy = $\dpi{80} \frac{1}{2} I\omega ^2$

=(1/2)*(1.82*10^-4)*(150.72)^2

=6.49 j