Calculating final rotational speed from angular velocity

AI Thread Summary
The discussion revolves around calculating final rotational speed from angular velocity, specifically transitioning from 1.5 rev/s to 9.4 rad/s. Participants clarify the relationship between moment of inertia (I) and angular velocity (w), emphasizing the conservation of angular momentum (L). It is established that if the final moment of inertia increases by a factor of three, the final angular velocity decreases correspondingly. The final angular velocity is calculated to be 3.1 rad/s based on this relationship. The calculations and concepts are confirmed as correct, providing clarity on the problem.
Anmol Dubey
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Homework Statement
An ice skater is spinning with a rotational speed of 1.5 rev/s. When he extends his arms and one leg, his rotational inertia increases by a factor of three. What is his final rotational speed?
Relevant Equations
Angular momentum is conserved
L = Iw
L (final) = L (initial)
I(initial)*w(initial) = I(final)*w(final)
I have no idea how to go about this. Any help would be appreciated thanks :)
Edit: I converted the 1.5 rev/s to rad/s = 9.4 rad/s
 
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What do you know about the quantities in your last relevant equation?
 
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Orodruin said:
What do you know about the quantities in your last relevant equation?
Like I = mr2?
w = Δθ/Δt
I didn't get what you mean by quantities
 
Anmol Dubey said:
Like I = mr2?
w = Δθ/Δt
I didn't get what you mean by quantities
No, what does the problem formulation tell you about these quantities:
I(final)
I(initial)
w(final)
w(initial)
 
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Orodruin said:
No, what does the problem formulation tell you about these quantities:
I(final)
I(initial)
w(final)
w(initial)
Since L is conserved
If I(final) is increased by a factor of 3, the w(final) is decreased by a factor of 3 so that L(final) = L(initial)
I(initial)*w(initial) = I(final)*w(final)
x*9.4 rad/s = 3x * w
so w(final) = 9.4 rad/s / 3
= 3.1 rad/s
Is that correct?
 
Anmol Dubey said:
Is that correct?
Yes.
 
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Orodruin said:
Yes.
Thank you for helping:biggrin:
 
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