# Angular acceleration of a falling arm

In summary: I'm excited to learn more about physics and problem solving.In summary, the conversation discusses the initial downward angular acceleration of a rod held out sideways and relaxed, and how to calculate the torque produced by gravity about the shoulder. The formula for torque is used and the value for the length in the formula for inertia is clarified. The center of mass and its effect on the torque is also discussed.

## Homework Statement

Assume your arm is a solid rod, 0.740 m in length, free to rotate about your shoulder (so its
moment of inertia is (1/3)ML^2). If you hold your arm out sideways, and then completely relax
your muscles, what is the initial downward angular acceleration of your arm?

τ=LF=Iα
F=mg
I=(1/3) ML^2

## The Attempt at a Solution

I'm not given a specific formula to use, but I think the above formula for Torque applies. I'm not using conservation of energy because I don't have a value for the height to use for potential energy.

If I plug the formulas for inertia and Fg into the torque formula:

Lmg=(1/3)ML^2 α

α=3g/L = (3*9.81)/0.740 = 39.77 but the answer key says the answer is 19.9 (half).What am I doing wrong here?

Last edited:

## Homework Statement

Assume your arm is a solid rod, 0.740 m in length, free to rotate about your shoulder (so its
moment of inertia is (1/3)ML^2). If you hold your arm out sideways, and then completely relax
your muscles, what is the initial downward angular acceleration of your arm?

τ=LF=Iα
F=mg
I=(1/3) ML^2

## The Attempt at a Solution

I'm not given a specific formula to use, but I think the above formula for Torque applies. I'm not using conservation of energy because I don't have a value for the height to use for potential energy.

If I plug the formulas for inertia and Fg into the torque formula:

Lmg=(1/3)ML^2 α

α=3g/L = (3*9.81)/0.740 = 39.77 but the answer key says the answer is 19.9 (half).

What am I doing wrong here?
What is the location of the center of mass of the arm ?

L/2, the very center of the length?

L/2, the very center of the length?
Yes.

That's where gravity effectively acts.

So, what is the torque produced by gravity about the shoulder ?

τ=(L/2)F=(L/2)mg
But the value of the length in the formula for inertia will remain the full length right??

τ=(L/2)F=(L/2)mg
But the value of the length in the formula for inertia will remain the full length right??
Right !

awesome! Thank you so much for the help!

awesome! Thank you so much for the help!
You're welcome.

By The Way, welcome to PF !

Thanks!

## 1. What is angular acceleration?

Angular acceleration is the rate of change of angular velocity of an object. It is a measure of how much an object's angular velocity changes in a given amount of time.

## 2. How is angular acceleration related to linear acceleration?

Angular acceleration and linear acceleration are related through the radius of rotation. The linear acceleration of an object is equal to the angular acceleration multiplied by the radius of rotation.

## 3. How is angular acceleration of a falling arm calculated?

The angular acceleration of a falling arm can be calculated by dividing the change in the arm's angular velocity by the time it takes to fall. This can be represented by the formula: angular acceleration = (final angular velocity - initial angular velocity) / time.

## 4. What factors affect the angular acceleration of a falling arm?

The main factors that affect the angular acceleration of a falling arm are the mass of the arm, the distance from the center of rotation, and the force of gravity. A larger mass or longer arm will result in a higher angular acceleration, while a smaller distance from the center of rotation or a stronger force of gravity will also increase the angular acceleration.

## 5. How does air resistance affect the angular acceleration of a falling arm?

Air resistance can affect the angular acceleration of a falling arm by slowing down its rate of descent. This is because air resistance creates a force in the opposite direction of the arm's motion, causing it to experience a net force that is less than the force of gravity. As a result, the arm will experience a smaller angular acceleration compared to if there was no air resistance present.

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