Recent content by ht9000

Awesome, thanks a lot for the help

ω in terms of T is (Δθ)/T right? 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?

Yeah, multiply by one hundred, and to get that expression for the change in T, would I just have to take 2π divided by the expression I have now?

Is this right? ω_f = (r_i)^2 * ω_i / (r_f)^2 then, (ω_f-ω_i)/ω_i = (ω_f/ω_i) - 1 Using the expression I got for ω_f above, that gives: ((r_i)^2 / (r_f)^2) -1

So it will be the change in the period divided by the original period.

You divide it by the original quantity and multiply by 100

Right now what I have is: (r_f)^2 * ω_f = (r_i)^2 * ω_i where _f indicates the final state and _i indicates the initial state. Can I turn that into some expression for the percent change?

Yeah, and then we know one of the ω's already because we know the period of the Earth right now, which is 23 hours 56 minutes 4.1 seconds, or 86164.1 s. Take 2π divided by that time and that is the initial ω, and using that I can solve for the one I need.

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?

Can we solve it with the rotational KE and its relation to PE?

Homework Statement The Earth's radius is 6371 km. If the Earth's radius were to increase by 30 m (0.03 km), but no change in mass, by what percentage would the Earth's rotational period increase? (Model the Earth as a uniform sphere) Homework Equations ∑Torque = Iα v = rω a = rα KE = (1/2)Iω^2...