A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 19.7 m / s at a distance of 0.470 m from the joint and the moment of inertia of the forearm is 0.550 kg·m2, what is the rotational kinetic energy of the forearm?
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To calculate the rotational kinetic energy of the forearm, we can use the formula:
Rotational Kinetic Energy = (1/2) * moment of inertia * angular velocity^2
First, we need to find the angular velocity. The linear velocity of the ball relative to the elbow joint can be related to the angular velocity by the following formula:
v = r * ω
Where:
- v is the linear velocity (19.7 m/s)
- r is the distance from the joint (0.470 m)
- ω is the angular velocity (unknown)
Rearranging the formula, we can solve for ω:
ω = v / r
Substituting the given values, we get:
ω = 19.7 m/s / 0.470 m
Calculating this, we find:
ω = 41.91 rad/s
Now that we have the angular velocity, we can calculate the rotational kinetic energy:
Rotational Kinetic Energy = (1/2) * moment of inertia * (angular velocity)^2
Substituting the given values, we get:
Rotational Kinetic Energy = (1/2) * 0.550 kg·m^2 * (41.91 rad/s)^2
Evaluating this equation, we find:
Rotational Kinetic Energy ≈ 475.06 J
Therefore, the rotational kinetic energy of the forearm is approximately 475.06 Joules.