In softball, the pitcher throws with the arm fully extended (straight at the elbow). in a fast pitch the ball leaves the hand with a speed of 139 km/h. find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder.
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Ответ:
K.E = 1490.73 J
Explanation:
Given:
Speed of the ball (V)= 139 km/h = 38.61 m/s
Moment of inertia, I = 0.720 kg-m²
Distance, r = 0.600 m
Angular velocity (ω) =
(ω) = =64.35 rad/sec
The kinetic energy (K.E) is given as
K.E =
substituting the values in the above equation we get
K.E =
K.E = 1490.73 J
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