A dart player holds a dart in her hand with her forearm initially in a vertical

Question

A dart player holds a dart in her hand with her forearm initially in a vertical position. She quickly accelerates the forearm forward and releases the dart, holding the upper arm fixed, as good dart players usually do. Find the torque that must be applied to the forearm by muscles connecting it to the upper arm in order for the dart to leave the hand with a speed of 11.5 m/s after the forearm has rotated through an angle of 0.302 rad. The radial distance from the dart to the pivot point at the elbow is 0.400 m. Treat the forearm as a thin rod of mass 2.00 kg and length 0.400 m.

Explanation / Answer

the speed of the dart leaving the hand is v = 15 m/s The angular displacement of the arm is ? = 0.316 rad The distance of the dart to pivot is R = 0.4 m mass of the arm is m = 2 kg length of the arm is L = 0.4 m The final angular velocity of the arm is ?f = v/R = (15 m/s)/(0.4 m) = 37.5 rad/s The equation of motion of the arm is ?f 2 - ?i2 = 2a? (37.5 rad/s)2 = 2a(0.316 rad) a = 2.22*103 rad/s2 The moment of inertia of the arm is I = mL2/3 = (2 kg)(0.4 m)2/3 = 0.107 kgm2 The torque required is t = Ia = (0.107 kgm2)(2.22*103 rad/s2) = 237.54 Nm

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