Rotational inertia, angular speed, and kinetic energy

In summary, two skaters of mass 50 kg each approach each other with opposite velocities of 1.7 m/s along parallel paths separated by 4.0 m. They grab onto opposite ends of a long pole with negligible mass and rotate around the center of the pole. The friction between the skates and ice is negligible. The radius of the circle they skate in is calculated in part (a) and the angular speed in part (b). In part (c), the kinetic energy of the two-skater system is calculated. In part (d), conservation of angular momentum is used to calculate the new angular speed as the skaters pull themselves closer together. Finally, in part (e), the kinetic energy of the system after
  • #1
bonekrushur
3
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.
 
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  • #2
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.
 
  • #3
Thanks. I got it.
 

FAQ: Rotational inertia, angular speed, and kinetic energy

What is rotational inertia and how is it different from mass?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in rotational motion. It is similar to mass in linear motion, but instead of measuring an object's resistance to changes in linear motion, it measures its resistance to changes in rotational motion. The greater the rotational inertia, the harder it is to rotate an object.

How does angular speed affect an object's kinetic energy?

Angular speed is the rate at which an object rotates around a fixed axis. It is directly proportional to an object's kinetic energy, meaning that as angular speed increases, so does the object's kinetic energy. This is because an object with a higher angular speed has a higher rotational velocity, resulting in more kinetic energy.

How do you calculate rotational inertia?

The formula for calculating rotational inertia is I = mr², where I is the rotational inertia, m is the mass of the object, and r is the distance between the object and its axis of rotation. This formula is similar to the formula for calculating mass (m = ρV), but instead of using volume, it uses the distance from the axis of rotation.

Can rotational inertia be changed?

Yes, rotational inertia can be changed by altering the mass or distribution of mass in an object. For example, if the mass of an object is concentrated closer to its axis of rotation, its rotational inertia will be lower. On the other hand, if the mass is distributed farther away from the axis of rotation, the rotational inertia will be higher.

What is the relationship between rotational inertia and torque?

Torque is the measure of the force that causes an object to rotate around an axis. The relationship between rotational inertia and torque can be described by the equation τ = Iα, where τ is torque, I is rotational inertia, and α is angular acceleration. This equation shows that the greater the rotational inertia, the harder it is to change the object's angular velocity, and therefore, the greater the torque needed to do so.

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