Calculating Angular Acceleration and Required Force for Forearm Throw

  • Thread starter Thread starter Amerk2
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating angular acceleration and the force required by the triceps muscle for a forearm throw involving a 1 kg ball. The user successfully calculated the angular acceleration as 78.341 rad/s² but is struggling with the required force, which the answer key states should be 670 N. The user applied the torque equation but is unsure how to incorporate the ball's mass into the moment of inertia, as they believe it should be treated as a point mass at the length of the forearm. Clarification is needed on how to properly factor in the ball's weight and its distance from the elbow joint in the calculations. Understanding these components is crucial for solving the problem accurately.
Amerk2
Messages
1
Reaction score
0
Hi,

I'm new here, and in dire need of help. This problem has been stumping me for several hours now, and I don't seem to be getting any closer to understanding or solving it.

The premise: Assume that a 1 kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle. The ball is accelerated uniformly from rest to 8.5 m/s in .35 seconds, at which point it is released.

A diagram is shown of a elbow, presumably in the 90 degree position.

A.) Find the angular acceleration.

B.) The force required of the triceps muscle.

I've got A, and I know it's correct at 78.341 rad/s^2.

Now B is supposed to be 670 N according to the answer key, but I'm not getting it.

My approach so far:

I(moment of inertia) * angular acceleration = torque
R(perpendicular - this is the length of the lever it seems) * Force = Torque

So,

I*angular a = RF

The real values I used were:

I = .118523 ( I used I = 1/3(ml^2) for that, the equation for the I of a long uniform rod with axis at one end). I used .31 m for its length, and 3.7 kg for its weight.

Angular acceleration was the one I supplied above for A, and R was .025 m (for the right side of the equation).

I'm not sure where I'm going wrong, but I know I'm missing the effect of the ball held in the hand. It weighs 1 kg, but I don't think I'm supposed to factor it into the moment of inertia, as no radius is given.

Any thoughts?
 
Last edited by a moderator:
Physics news on Phys.org
Where's the diagram?
 
Amerk2 said:
I'm not sure where I'm going wrong, but I know I'm missing the effect of the ball held in the hand. It weighs 1 kg, but I don't think I'm supposed to factor it into the moment of inertia, as no radius is given.
Treat the ball as a point mass. Its distance from the axis will be the length of the forearm.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top