Stuck on physics questions. In descriptin box below.A baseball pitchers foremarm (mass of 1.50kg and length to hand centre 0.350m) in the process of pitching a ball mass of (mass of 1.50kg from...

Stuck on physics questions. In descriptin box below.

A baseball pitchers foremarm (mass of 1.50kg and length to hand centre 0.350m) in the process of pitching a ball mass of (mass of 1.50kg from rest). If the toal torque acting on the forearm (includes hand) and ball (combined) about the elbow joint is 22.8N.m and this torque acts for 0.15 sec, approximate the realse of the speed of the ball. Assume a simple geometrically- segemented anthropometric biomechnaical model.

From my calculations I got 20m/s not sure if I am correct. Would be nice to how you solve this.

Asked on by ledjic

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

The moment of inertia of the forearm (modeled by a rod rotating about one end) and the ball in hand (modelled by a single mass at distance R) rotating about the elbow is

`I = I_("forearm") +I_("ball") = (M_1*L^2)/3 +M_2*L^2 =`

`=(1.5*0.35^2)/3 +1.5*0.35^2 =0.245 kg*m^2`

The angular acceleration produced by the torque applied is

`T =I*epsilon`

`epsilon = T/I =22.8/0.245 =93.06 s^-2`

The linear acceleration at the hand tip is

`a =epsilon*L = 93.06*0.35 =32.57 m/s^2`

Therefore the release speed of the ball from the hand is:

`v = a*t =32.57*0.15 =4.88 m/s`

(The speed is 20 m/s only if calculating the momentum of inertia one does not take into account the momentum of inertia of the ball itself).

Answer: The speed of the ball is 4.88 m/s

Sources:

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