Torque in throwing a ball. A baseball player extends his arm straight upward to

Question

Torque in throwing a ball.
A baseball player extends his arm straight upward to catch a 0.143 kg batted baseball moving horizontally at 43.4 m/s. It's 64.1 cm from the player's shoulder joint to the point where the ball strikes his hand, and his arm remains stiff while it rotates about the shoulder joint during the catch. The player's hand recoils horizontally a distance of 5.30 cm while he stops the ball with constant acceleration.


What torque does the player's arm exert on the ball?
T=?

***HINT***
In this problem you need to find the average force that the player uses to stop the ball. I would recommend using energy considerations. The work done to stop the ball equals the initial kinetic energy of the ball. Once you have the average force, you can find the torque around the player's shoulder.
***

Explanation / Answer

Mass of the ball, m = 0.143 Kg Horizontal speed of the ball, u = 43.4 m/s Distance between the players shoulder and line of motion of the ball, r = 64.1 cm = 0.641 m Recoil distance S = 5.30 cm = 0.053 m From Work energy thorem                                  Work done W = Change in Kinetic energy           Average force F * Distance S = (1/2)mu^2 - 0                                      F*(0.053 m) = 0.5 * (0.143 Kg)*(43.3 m/s)^2                                Average force F = 2529.33 N Hence Torque applied = Fr                                        = (2529.33 N)(0.641 m)                                        = 1621.3 Nm                                                                   

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