Angular Momentum of a playground ride

In summary, a playground ride consists of a disk of mass M = 46 kg and radius R = 2.0 m mounted on a low-friction axle. A child of mass m = 19 kg runs at speed v = 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk. The angular speed is 0.502299 m/s.
  • #1
HasuChObe
31
0
A playground ride consists of a disk of mass M = 46 kg and radius R = 2.0 m mounted on a low-friction axle. A child of mass m = 19 kg runs at speed v = 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.

What is the angular speed?

I found the total angular momentum and used L_rot = I*w. Didn't work =/ Since the disk does not move, does that mean that L_total = L_rot? Or am I not getting something about L_trans? Cuz I assumed L_trans is zero because the system isn't moving.
 
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  • #2
Translational momentum is not conserved, since the axle of the disk is fixed. But angular momentum is conserved. What's the initial angular momentum of the child just as she jumps onto the disk? What's the total rotational inertia of the system (disk + child) after she jumps on? Use that to solve for the angular speed.
 
  • #3
Can you explain how I should be adding up the rotational inertias? Not sure what the end result should look like.
 
  • #4
The total rotational inertia of the system is just the sum of the individual rotational inertias of the parts:
(1) the disk (use a standard formula to calculate its I)
(2) the child (treat the child as a point mass a distance R from the center)
 
  • #5
Hint: Assuming the child is a point mass (unless you want to integrate over the child :biggrin: ) you can use the parrallel axis theorem which says that the I of the child is [tex]mr^2[/tex]. Add this to the I of the disk to find total I. You add I's just like you would add m's, nothing complicated there. Use that to find total [tex]\omega[/tex]
 
  • #6
So L_total would be 2*19*2.3 -> 87.4 kg m/s. And I would be 19*2^2 + .5*49*2^2. L_total / I should be my answer then? I get 0.502299. Thnx for the help btw =]
 

Related to Angular Momentum of a playground ride

1. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object. It is the product of an object's moment of inertia and its angular velocity.

2. How is angular momentum related to playground rides?

Playground rides, such as a merry-go-round or a swing, involve rotational motion. As the ride spins or swings, it has an angular velocity and therefore angular momentum.

3. How does the angular momentum of a playground ride affect its motion?

The angular momentum of a playground ride is conserved, meaning it stays constant unless acted upon by an external force. This affects the motion of the ride, as any changes in angular momentum will result in changes in rotational speed or direction.

4. How is angular momentum calculated for a playground ride?

The angular momentum of a playground ride can be calculated by multiplying the moment of inertia, which is determined by the mass and distribution of mass of the ride, by the angular velocity, or speed of rotation.

5. How can the angular momentum of a playground ride be changed?

The angular momentum of a playground ride can be changed by applying an external torque, such as pushing or pulling on the ride, or by changing the distribution of mass on the ride. These changes will result in changes in rotational speed or direction for the ride.

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