Calculating Yo-Yo Angular Momentum on an Inclined Plane

Click For Summary
SUMMARY

The discussion focuses on calculating the angular momentum of a yo-yo as it leaves the hand, which acts as an inclined plane. The yo-yo has a mass of 0.2 kg and is released from an incline of approximately 50 degrees. The gravitational force acting on the yo-yo is 1.96 N. Two scenarios are presented: one without friction, where the acceleration down the incline is calculated using the formula mg sin(θ), and another with friction, which complicates the angular momentum due to the rotational inertia and radius of the yo-yo. The unique design of the yo-yo ensures that it rotates as it descends, making it distinct from other disks.

PREREQUISITES
  • Understanding of basic physics concepts such as angular momentum and acceleration.
  • Familiarity with inclined plane dynamics and free body diagrams.
  • Knowledge of gravitational force calculations (mg).
  • Basic principles of rotational motion and inertia.
NEXT STEPS
  • Study the equations of motion for inclined planes in physics.
  • Learn about the relationship between linear velocity and angular velocity in rotating objects.
  • Explore the effects of friction on angular momentum and rotational motion.
  • Investigate the principles of energy conservation in rotational systems.
USEFUL FOR

Students engaged in physics projects, particularly those focusing on mechanics, as well as educators teaching concepts related to angular momentum and inclined planes.

Mr. Snookums
Messages
20
Reaction score
0
Physics project on the yo-yo. It's for bonus marks.

I need to calculate the angular momentum of a yo-yo when it leaves your hand. The hand represents an inclined plane which imparts the spin. I'm keeping it simple and won't involve friction.

What I'm trying to get at is the equation for acceleration down an inclined plane. I've drawn a free body diagram but I just can't get anywhere on it. Yo-yo's mass is 0.2kg, and the incline would be about 50 degrees, I guess.

mg for the yo-yo is 1.96N, theta is 50 degrees. How do I get the acceleration down the inclined plane?

We haven't covered inclined planes and angular momentum yet, it's not even in the curriculum. That's why it's a bonus project.
 
Physics news on Phys.org
Does this problem involve friction?

Note the two cases

Case 1: No friction
If there is no friction, the yo yo carries no angular momentum. The acceleration is simply [tex]mg sin\theta[/tex]

Case 2: Friction
This makes things a little bit more complicated. Friction actually create angular momentum, and it also create a moment about the yo-yo. In this case you will need to express friction in terms of the rotational inertia and the radius of the yo-yo.

Good luck
 
Mr. Snookums said:
Physics project on the yo-yo. It's for bonus marks.

I need to calculate the angular momentum of a yo-yo when it leaves your hand. The hand represents an inclined plane which imparts the spin. I'm keeping it simple and won't involve friction.

What I'm trying to get at is the equation for acceleration down an inclined plane. I've drawn a free body diagram but I just can't get anywhere on it. Yo-yo's mass is 0.2kg, and the incline would be about 50 degrees, I guess.

mg for the yo-yo is 1.96N, theta is 50 degrees. How do I get the acceleration down the inclined plane?

We haven't covered inclined planes and angular momentum yet, it's not even in the curriculum. That's why it's a bonus project.

This would make sense if you were talking about a disk other than a yo-yo. The post by Hyperreality points out that without friction there would be no angular momentum without friction. That would be true for any old disk, and friction would be needed to impart a rotation. But a yo-yo is a different thing. The string wound around the axle makes it impossible for the yo-yo to move without beginning to rotate. If you know the radius of the axle, and you give the yo-yo an initial velocity downward, you can determine the rate of rotation in terms of the velocity and the radius. You can also calculate the change in both velocity and angular velocity (and change in momentum and angular momentum) resulting from gravity acting on the yo-yo. The fact that so much of the yo-yo energy is rotational energy is what makes the yo-yo climb back up the string to your hand. If your yo-yo is a sleeper, and you throw it hard enough there will still be plenty of energy left to climb the string even after friction slows the rotation at the bottom.
 

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
5K
  • · Replies 49 ·
2
Replies
49
Views
10K
Replies
2
Views
1K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
7
Views
2K
Replies
11
Views
2K
  • · Replies 18 ·
Replies
18
Views
6K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K