Answer:
Rotational kinetic energy = 0.099 J
Translational kinetic energy = 200 J
The moment of inertia of a solid sphere is [tex]I = \frac{2}{5}mr^2[/tex].
Explanation:
Rotational kinetic energy is given by
[tex]\text{RKE} = \frac{1}{2}I\omega^2[/tex]
where I is the moment of inertia and ω is the angular speed.
For a solid sphere,
[tex]I = \frac{2}{5}mr^2[/tex]
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
[tex]\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2[/tex]
[tex]\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}[/tex]
Translational kinetic energy is given by
[tex]\text{TKE} = \frac{1}{2} mv^2[/tex]
where v is the linear speed.
[tex]\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}[/tex]
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
