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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 22 m/s at a distance of 0.416 m from the joint and the moment of inertia of the forearm is 0.524 kg m2 , what is the rotational kinetic energy of the forearm?

Answer :

boffeemadrid

Answer:

732.75702 Joules

Explanation:

v = velocity of the ball relative to the elbow joint = 22 m/s

r = Distance from the joint = 0.416 m

I = Moment of inertia of the forearm = 0.524 kg m²

Angular speed is given by

[tex]\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{22}{0.416}\ rad/s[/tex]

The rotational kinetic energy of the ball is

[tex]K=\frac{1}{2}I\omega^2\\\Rightarrow K=\frac{1}{2}\times 0.524\times \left(\frac{22}{0.416}\right)^2\\\Rightarrow K=732.75702\ J[/tex]

The rotational kinetic energy of the forearm is 732.75702 Joules

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