rotational kinetic energy?

Question

An ice skater is spinning in place(doing pirouette) and increases his rate of rotation by a factor of 2 by pulling his arms in against his body and reducing his moment of inertia. Which of the following statements correctly describes his rotational kinetic energy?

a) his rotational kinetic energy is zero
b) his rotational kinetic energy remains constant and not zero.
c) his rotational kinetic energy increases by a factor of 2
d) his rotational kinetic energy increases by a factor of 4
e) none of the above.

Answer 1

Rotational kinetic energy is given by E0= I*w*w/2
Where I is the moment of inertia and w is the angular velocity.

As there is no external torque acting whilst he moves his hands inwards the angular momentum is a constant. This means that if the angular velocity has doubled then the moment of inertia must have halved to keep the angular momentum the same. This is because angular momentum is expressed as the product of the moment of inertia and the angular velocity.

Finding the new rotational kinetic energy, by halving the moment of inertia and doubling the angular velocity gives:

E1 = ( I/2 * 2w * 2w )/ 2
= 2 * E0

So c) is correct.

Answer 2

a), b), d) are definitely wrong.

c) is not quite right (but might be what the test writer thought was the right answer)

Here's the scoop, in two cases:
Suppose the ice-skater is a real dum-dum. I mean, of course, that we model him as being two equal masses (m) connected by a weightless rod of length L.
Then initially, he's rotating at angular frequency w, so the speed of each of the masses is w*L/2, and their individual kinetic energies are (m/2)(wL/2)^2 = mL^2w^2/8, so the total kinetic energy is twice that: m(wL)^2/4.
The angular momentum is: 2*m*(wL/2)(L/2) = mwL^2/2.

Then he pulls the masses together, and w => 2w. Angular momentum is conserved, so L => L/sqrt(2). Therefore, the new kinetic energy is: m(2wL/sqrt(2))^2/4 = m(wL)^2/2, which is twice what it was before.

So why isn't c) the right answer? Because the ice-skater isn't really a dum-dum: he has an inner core that is not compressed, and this has a "moment of inertia", I, that is unchanged. The real expression for his kinetic energy and angular momentum are:

Total AM = mwL^2 + Iw
Total KE = m(wL)^2/4 + Iw^2/2

So when w => 2w
conservation of AM implies:
(mL'^2 + I)(2w) = (mL^2 + I)w
2mL'^2 = mL^2 - I
L' ^2= (mL^2 - I)/2m

So new total KE is:
m(2wL')^2/4 + I(2w)^2/2
= 2*Iw^2 + (mw^2(L'^2)
= 2*Iw^2 + (mw^2)(mL^2 - I)/(2m)
= 2*Iw^2 +(w^2/2)(mL^2 - I)

The old total KE was:
m(wL)^2/4 + Iw^2/2
= Iw^2/2 + (w^2/4)(mL^2)

So the ratio of New to Old KE is:
[2*Iw^2 +(w^2/2)(mL^2 - I)]/[Iw^2/2 + (w^2/4)(mL^2)]
= [2*I + (1/2)(mL^2 - I)][I/(2) + (1/4)(mL^2)]
= [8*I + 2(mL^2 - I)]/[2*I + (mL^2)]
= 2 * [3*I + mL^2]/[2*I + (mL^2)]

So you see that if I = 0 (the skater is a dum-dum), the ratio of KEs is =2; but if the skater has any fixed core section that doesn't compress inward (I not = 0), the KE increase by a little more than that factor of 2.

Answer 3

If something has mass and rotation, the KE is not zero, so a) is out.

In order for KE to increase, energy must be added to the system by an outside source, so c) and d) are out.

As no energy is added to the system, if one neglects the air friction and the friction of the skate on the ice, then KE would remain constant an b) would be valid.

Answer 4

"d)" is acceptable.
the formula is k = 1/2*i*(w^2)
"i" is rotational inertia and w is arate of rotation .
so it will be 4 times more.

Answer 5

f) you