The discussion focuses on calculating the percentage increase in the Earth's rotational period when its radius increases by 30 meters while maintaining constant mass. The key equations utilized include the moment of inertia for a sphere, \(I = \frac{2}{5}mr^2\), and the conservation of angular momentum, \(L = I\omega\). The final expression derived for the percentage change in the period \(T\) is \(\frac{(r_f)^2}{(r_i)^2} - 1\), where \(r_f\) and \(r_i\) represent the final and initial radii, respectively. Participants emphasize the importance of using angular momentum to relate changes in rotational speed and period.
PREREQUISITESPhysics students, educators, and anyone interested in understanding the dynamics of rotating bodies and the implications of changes in mass distribution on rotational periods.
Can we solve it with the rotational KE and its relation to PE?TSny said:Welcome to PF!
##\sum \tau = I \alpha## only applies to a rigid body rotating about a fixed axis. Here, the Earth is not acting as a rigid body.
Can you relate this problem to any other examples you've been studying lately where a rotating system changes its mass distribution while no external torque acts?
Rather than energy, there's a much more appropriate physical quantity that can be used. Have you read or discussed in class what goes on when an ice skater pulls in her arms in order to spin faster?ht9000 said:Can we solve it with the rotational KE and its relation to PE?
Yes. (Distinguish between the two ω's.)ht9000 said:Oh yeah! Conservation of angular momentum! Which is L = Iω. And it will be the same at its original size and after it grows a little bit.
(2/5)mr^2 ω = (2/5)m(r+.03)^2 ω, m's cancel out and I solve for ω?
Right now what I have is:TSny said:OK. But since you are asked to find the % change in ω, you could first work out an expression for the % change and simplify the expression before substituting numbers. That way, you might not even need to know the initial rate of spin.
You divide it by the original quantity and multiply by 100TSny said:Yes. How to you find the % change in a quantity?
So it will be the change in the period divided by the original period.TSny said:
Is this right?TSny said:It might be helpful to note that ##\frac{T_f - T_i}{T_i} = \frac{T_f}{T_i}-1##.
No. I would go back to the angular momentum relation and express it in terms of T rather than ω. Then you can easily find an expression for Tf/Ti.ht9000 said:to get that expression for the change in T, would I just have to take 2π divided by the expression I have now?
ω in terms of T is (Δθ)/T right? I substituted that in place of ω and when I solved for it, it gave meTSny said:No. I would go back to the angular momentum relation and express it in terms of T rather than ω. Then you can easily find an expression for Tf/Ti.
Yes, with Δθ = 2##\pi##.ht9000 said:ω in terms of T is (Δθ)/T right?
Yes, that's it.I substituted that in place of ω and when I solved for it, it gave me
((r_f)^2 / (r_i)^2) - 1
Is that the expression for % change in the period?
Awesome, thanks a lot for the helpTSny said:Yes, that's it.