Rotational inertia, angular speed, and kinetic energy

Click For Summary
SUMMARY

The discussion centers on a physics problem involving two skaters with a mass of 50 kg each, who rotate around a pole after grabbing it while skating towards each other at 1.7 m/s. The key equations used include angular speed (ω = v/r) and kinetic energy (KE = 0.5 * I * ω²). The solution to part (d) requires applying the conservation of angular momentum, as external torques are absent. The kinetic energy changes due to the work done by the skaters when pulling themselves closer together, which adds energy to the system.

PREREQUISITES
  • Understanding of rotational dynamics and angular momentum
  • Familiarity with the equations of motion for rotating bodies
  • Knowledge of kinetic energy calculations in rotational systems
  • Basic principles of conservation laws in physics
NEXT STEPS
  • Study the conservation of angular momentum in closed systems
  • Learn about the relationship between linear and angular velocity
  • Explore the concept of work done in mechanical systems
  • Investigate the effects of changing radius on angular speed and kinetic energy
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for practical examples of angular momentum and energy conservation principles.

bonekrushur
Messages
3
Reaction score
0

Homework Statement



Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 4.0 m. They have opposite velocities of 1.7 m/s each. One skater carries one end of a long pole with negligible mass, and the second skater grabs the other end of it as she passes. The skaters then rotate around the center of the pole. Assume that the friction between skates and ice is negligible.

(a) What is the radius of the circle they now skate in?
(b) What is the angular speed of the skaters?
(c) What is the kinetic energy of the two-skater system?
(d) Next, the skaters pull along the pole until they are separated by 0.7 m. What is their angular speed then?
(e) Calculate the kinetic energy of the system now.


Homework Equations



w= v/r
ke= 0.5*I*w^2


The Attempt at a Solution



I have figured out parts (a), (b), and (c). For part (d), I used the same process as in part (b). In (d), I assumed that the new radius was .35 m, and the velocity was still 1.7. So, 1.7/.35 should equal the new angular speed, but that answer does not work. In part (e), I still used what I think is the new radius, .35 m, and put the values into the Kinetic energy equation: 0.5*(50*(.35)^2+50*(.35)^2)*(1.7/.35)^2 = 0.5*I*w^2. This answer does not work either, but I have to be missing something in part (d) to begin with, and I have to get that first anyway to solve (e). Does anyone see what I'm forgetting here? Thanks.
 
Physics news on Phys.org
bonekrushur said:
In (d), I assumed that the new radius was .35 m,

So far so good. o:)

and the velocity was still 1.7.

I think that's where you went wrong. :frown:

Use conservation of angular momentum for part (d). The only things acting on the system are the skaters themselves, thus there are no external torques, which means that conservation of angular momentum applies. Set the angular momentum, before and after the skaters pull themselves closer together, equal to each other. Solve for the new [tex]\omega[/tex].

Why does the kinetic energy change? The skaters have to do work to pull themselves inwards. The skaters are adding energy to the system in the form of work. But change in kinetic energy or not, angular momentum is still conserved.
 
Thanks. I got it.
 

Similar threads

Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K