Conservation of Angular Momentum Question

AI Thread Summary
The discussion centers on the conservation of angular momentum in the context of an ice skater performing a toe spin. When the skater tucks in her arms, her moment of inertia decreases by 7.5%, resulting in an increase in angular velocity to 4.32 rad/s. The calculations confirm that the initial angular momentum equals the final angular momentum, validating the new angular velocity. Additionally, the change in kinetic energy is shown to be zero, as the initial and final kinetic energies are equal, demonstrating that kinetic energy remains constant when no external torque is applied. Overall, the skater's motion illustrates the principles of angular momentum conservation and its effects on rotational dynamics.
timtng
Messages
25
Reaction score
0
An ice skater doing a toe spin with outstretched arms has an angular velocity of 4 rad/s. She then tucks in her arms, decreasing her moment of inertia by 7.5%

a. What is the resulting angular velocity?
b. By what factor does the skater's kinetic energy change?

For a, I use IW = I'W' >> 1(4rad/s) = (1-.075)W', then solve for W'
I got 4.32 rad/s for W'. I don't know if I'm doing it correctly. Also, I need help on part b.

Thx
 
Physics news on Phys.org
Part a looks ok.

IIRC the formula for rotational energy is
1/2Iω2

The ratio is the old kinetic energy divided by the new kinetic energy.
 
.

a. The resulting angular velocity can be calculated using the conservation of angular momentum equation, which states that the initial angular momentum is equal to the final angular momentum. In this case, the initial angular momentum is given by I1W1, where I1 is the initial moment of inertia and W1 is the initial angular velocity. The final angular momentum is given by I2W2, where I2 is the final moment of inertia and W2 is the final angular velocity. Since the angular momentum is conserved, we can set these two equations equal to each other:

I1W1 = I2W2

We are given that the initial angular velocity is 4 rad/s and the moment of inertia decreases by 7.5% when the skater tucks in her arms. This means that the final moment of inertia is 0.925 times the initial moment of inertia. Therefore, we can rewrite the conservation of angular momentum equation as:

I1(4 rad/s) = (0.925I1)W2

Solving for W2, we get:

W2 = (4 rad/s)(I1/0.925I1) = 4.32 rad/s

Therefore, the resulting angular velocity is 4.32 rad/s.

b. To calculate the change in kinetic energy, we can use the equation for kinetic energy in terms of angular momentum, which is given by:

K = 1/2I1W1^2

The initial kinetic energy is given by:

K1 = 1/2I1(4 rad/s)^2 = 8I1

Similarly, the final kinetic energy is given by:

K2 = 1/2(0.925I1)(4.32 rad/s)^2 = 8I1

Therefore, the change in kinetic energy is:

ΔK = K2 - K1 = 8I1 - 8I1 = 0

This means that the skater's kinetic energy does not change when she tucks in her arms. This is due to the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless an external torque is applied. In this case, since no external torque is applied, the skater's angular momentum remains constant and therefore, her kinetic energy also remains constant.
 
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 .
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...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top