Question

A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so that it moves with an angular speed of 49 rad/s and a linear speed of 52 m/s. Assuming the baseball to be a uniform solid sphere, determine the rotational and translational kinetic energies of the ball in joules. KErotational = Incorrect: Your answer is incorrect. What is the moment of inertia of a solid sphere? J KEtranslational = J Additional Materials

Answer 1

Answer:

KE Rotational = 0.0986 J

KE Translational = 202.8 J

Explanation:

The rotational kinetic energy is given by:

KE Rotational = (1/2)Iω²

Where;

I = moment of inertia of a uniform solid sphere = (2/5)mr²

m = mass of sphere = 0.15 kg

r = radius of sphere = 0.037 m

ω = angular velocity (spin rate) = 49 rads/sec

Thus, KE Rotational = (1/2)(2/5)(0.15 x 0.037²)(49²) = 0.0986 J

The translational kinetic energy is:

KE Translational = (1/2)mv²

m = ball's mass = 0.15 kg

v = ball's velocity = 52 m/s

KE Translational = (1/2)(0.15)(52²) = 202.8 J

Answer 2

Answer:

Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$.

Explanation:

Rotational kinetic energy is given by

$\text{RKE} = \frac{1}{2}I\omega^2$

where I is the moment of inertia and ω is the angular speed.

For a solid sphere,

$I = \frac{2}{5}mr^2$

where m is its mass and r is its radius.

From the question,

ω = 49 rad/s

m = 0.15 kg

r = 3.7 cm = 0.037 m

$\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2$

$\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}$

Translational kinetic energy is given by

$\text{TKE} = \frac{1}{2} mv^2$

where v is the linear speed.

$\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}$