Moment of Inertia/ Kinetic Energy of ice skater

AI Thread Summary
An ice skater's moment of inertia decreases when she pulls her arms in, which affects her kinetic energy. The discussion centers on how her angular momentum is conserved, leading to an increase in angular speed. The kinetic energy equation shows that while her moment of inertia halves, her angular speed doubles, resulting in her final kinetic energy being twice her initial kinetic energy. This raises the question of energy conservation, as the increase in kinetic energy must be accounted for. The conclusion emphasizes the importance of understanding the relationship between moment of inertia, angular speed, and energy conservation in rotational dynamics.
klopez
Messages
21
Reaction score
0
An ice skater starts a spin with her arms stretched out to the sides. She balances on the tip of one skate to turn without friction. She then pulls her arms in so that her moment of inertia decreases by a factor of two. In the process of her doing so, what happens to her kinetic energy?
-It undergoes a change by an amount that obviously depends on how fast the skater pulls her arms in.
-It decreases by a factor of two.
-It is zero because her center of mass is stationary.
-It increases by a factor of four.
-It decreases by a factor of four.
-It remains constant.
-It increases by a factor of two.

I know that Ki = (1/2)Iω2 , so isn't her Kf = (1/2)(I/2)(2ω2) = (1/2)Iω2 = Ki ?
Doesn't the angular speed increase by a factor of two as well?
My guess would be "It remains constant" (And I can only choose one)

Can anyone give me advice on this problem? Thank you

Kevin
 
Physics news on Phys.org
The KE = (1/2) I \omega^2. You know how I changes, but how does \omega change?

Hint: What's conserved?
 
Tell me if this makes sense...

Conservation of angular momentum Li = Lf

wi = omega_i
wf = omega_fLi = I*wi
Lf = I/2 * wf

2wi = wf

Now in terms of kinetic energy

Ki = (1/2)I*wi^2

Kf = (1/2)*(I/2)*wf^2 (I plug in my wf from above)
Kf = (I/4)(4wi^2)
Kf = I*wi^2

When I compare both Ki and Kf now, Kf is twice as large as Ki because it does not have a (1/2) in its equation like Ki does.

Is this correct?
 
Perfect!
 
You seem to have got it right...but before you continue on...remember the conservation of energy principle. If Kf is twice as large as Ki, where did that additional energy come from??
 
Last edited:
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 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top