# How Does the String's Path Affect Angular Momentum and Final Speed?

• csnsc14320
In summary, the mass is conserved in both cases, but the final speed is different depending on how the mass is attached to the post.
csnsc14320

## Homework Statement

Mass m is attached to a post of radius R by a string. Initially it is a distance r from the center of the post and is moving tangentially with speed $$v_o$$. There are two cases: In case (a) the string passes through a hole in the center of the post at the top, while the string is gradually shortened by drawing it through the hole, and case (b) the string wraps around the outside of the post.

What quantities are conserved in each case? Find the final speed of the mass when it hits the post for each case.

## The Attempt at a Solution

So, for case (a):
Angular momentum is conserved: $$I_o \omega_o = I_f \omega_f$$

$$m R^2 \frac{v_o^2}{(R)^2} = m r^2 \frac{v_f^2}{r^2}$$

Solving for $$v_f$$ we should get the right answer.

Now, case B is where I'm stumped. My initial thought is that angular momentum is conserved again, but then I end up with the exact same case I have in (a). Does the fact that the mass is rotating about the circumference of the rod, as opposed to the center of the rod, affect the speed of the mass? And, if so, how would I get started on how to find that?

Solve for vf and see what you get. Does this make sense?

chrisk said:
Solve for vf and see what you get. Does this make sense?
Oops.

$$I_o \omega_o = I_f \omega_f$$

$$m R^2 \frac{v_o}{R} = m r^2 \frac{v_f}{r}$$

$$R v_o = r v_f$$

$$v_f = \frac{R}{r} v_o$$

Didn't mean to square my angular velocity in my first post.

So, knowing this for case (a), how do we obtain $$v_f$$ for case (b)?

The $$r_o = R$$ and $$r_f = r$$, and $$v_o$$ is the same, why wouldn't the answer be the same as in case a?

Unless angular momentum isn't conserved? But I see no torque provided to cause a change in angular momentum in the first place.

How is the length of the string changing when pulled through the center compared to wrapping around a pole with a radius of "a"?

## 1. What is conservation of angular momentum?

Conservation of angular momentum is a physical law that states that the total angular momentum of a system remains constant unless acted upon by an external torque. Angular momentum is a measure of an object's rotational motion and is calculated by multiplying its moment of inertia (resistance to change in rotation) by its angular velocity (rate of rotation).

## 2. How does conservation of angular momentum apply to everyday life?

Angular momentum conservation is observed in many everyday activities, such as ice skating, figure skating, and roller skating. When a skater spins, their angular momentum remains constant unless an external force, like friction, is applied. This is why skaters can spin faster by pulling in their arms, reducing their moment of inertia, and increasing their angular velocity.

## 3. Can angular momentum be transferred between objects?

Yes, angular momentum can be transferred between objects in a system. For example, when a skater throws a ball while spinning, the ball will inherit some of the skater's angular momentum, causing it to rotate as well. However, the total angular momentum of the system will remain constant.

## 4. How does conservation of angular momentum apply in space?

Conservation of angular momentum is essential in understanding the motion of objects in space. For example, when a planet orbits around a star, its angular momentum remains constant, causing it to maintain a constant distance and speed. This is also observed in the rotation of planets and moons.

## 5. What happens if there is an external torque acting on a system?

If an external torque is applied to a system, it will cause a change in the system's angular momentum. This can result in a change in the object's rotational speed, direction, or both. However, the total angular momentum of the system will still remain constant, as long as there are no other external forces acting on it.

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