Angular Velocity of Ice Skater Around a Pole

[Gwozdzilla]

Homework Statement

An Ice skater of mass m = 70kg is initially traveling at a speed v = 4 m/s along a straight path that brings his center of mass to within a distance b = .6m of a vertical pole fixed in the ice. He reaches out with his hand as he passes the pole and hangs on so that he pulls himself into a circular path of radius r = .8m (measured from the axis of rotation to his center of mass) around the pole. If the skater's moment of inertia about his center of mass is I = 1.40 kgm² , what is his final angular velocity ω about the pole, assuming that no torques act during this maneuver?

Homework Equations

L_i = L_f
I_iω_i = I_fω_f
or maybe...
L = mvr = Iω
or possibly...
mv_ir_i = mv_fr_f

or if energy is conserved...
.5mv² = .5Iω²

I'm mostly confused about which formula to use.

The Attempt at a Solution

mvr_i = Iω
(70)(4)(.6) = (1.4)ω
ω = 120 rad/s
But I thought that this was a relatively unreasonable speed for the skater to be going, so I tried this...

.5(70)(4)² = .5(1.4)ω²
ω = 28.3 rad/s

But I don't think energy is conserved, so I also tried this...

mvr = mvr
(70)(4)(.6) = (70)(v)(.8)
v = 2.14 m/s
v=ωr
3 = ω(.8)
ω = 3.75 rad/s

This seemed reasonable to me, but I didn't use the moment of inertia, and I have nothing to check my work against. How do I know which formula is the correct one to use in this situation?

 

[haruspex]

mvr_i = Iω
(70)(4)(.6) = (1.4)ω

That I is the MoI about the skater's centre of mass, but the rotation is about the pole. How do you correct for that?

 

[Gwozdzilla]

I = I_cm + Md²
I = (1.4) + (70)(.8)²
I = 46.2 kgm²

mvr = Iω

(70)(4)(.6) = (46.2)ω
3.64 rad/s = ω

Would this be the correct answer?

 

[haruspex]

I = I_cm + Md²
I = (1.4) + (70)(.8)²
I = 46.2 kgm²

mvr = Iω

(70)(4)(.6) = (46.2)ω
3.64 rad/s = ω

Would this be the correct answer?

Looks right to me.

 

[Nickie]

As a scientist, it is important to carefully consider which formula to use in a given situation. In this case, we are dealing with angular velocity, which is the rate of change of angular displacement. Therefore, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

In this problem, we are given the initial angular velocity (zero), the moment of inertia, and the final angular velocity (what we are looking for). Therefore, we can use the formula Iiωi = Ifωf, where Ii and ωi represent the initial values and If and ωf represent the final values.

Plugging in the values, we get:

(1.40 kgm2)(0 rad/s) = (1.40 kgm2)(ωf)
ωf = 0 rad/s

This means that the skater's final angular velocity around the pole is zero, which makes sense because the skater is no longer rotating around the pole once he grabs onto it.

It is also important to note that the skater's initial angular momentum (L = Iiωi) is equal to his final angular momentum (L = Ifωf) since no external torques act on the system. This is known as the conservation of angular momentum.

In summary, the correct formula to use in this situation is Iiωi = Ifωf, and the skater's final angular velocity around the pole is 0 rad/s. I hope this helps clarify the confusion and demonstrates the importance of carefully selecting the appropriate formula in scientific problem-solving.