Rotational period for constant mass but different volume

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Homework Help Overview

The problem involves calculating the percentage change in the Earth's rotational period if its radius increases while maintaining the same mass. The context is rooted in rotational dynamics, specifically modeling the Earth as a uniform sphere.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the moment of inertia and conservation of angular momentum in relation to the problem. There are questions about the relevance of rigid body dynamics and comparisons to other rotating systems. Some participants explore the relationship between rotational kinetic energy and potential energy.

Discussion Status

The discussion is active with participants exploring various approaches to derive expressions for the percentage change in rotational period. Guidance has been offered regarding the use of angular momentum and the need to express quantities in terms of the period rather than angular velocity.

Contextual Notes

Participants are navigating the constraints of the problem, including the assumption of constant mass and the implications of changing the radius. There is a focus on deriving expressions without substituting numerical values initially.

ht9000
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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

The Attempt at a Solution


Moment of inertia for a sphere is (2/5)mr^2, so ∑Torque = (2/5)mr^2 * (a/r)
 
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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?
 
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?
Can we solve it with the rotational KE and its relation to PE?
 
ht9000 said:
Can we solve it with the rotational KE and its relation to PE?
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?
 
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?
 
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?
Yes. (Distinguish between the two ω's.)
 
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.
 
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.
 
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.
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?
 
  • #10
Yes. How to you find the % change in a quantity?
 
  • #11
TSny said:
Yes. How to you find the % change in a quantity?
You divide it by the original quantity and multiply by 100
 
  • #14
Yes. (Sorry, somehow I was thinking the question asked for percent change in ω rather than T.)
 
  • #15
It might be helpful to note that ##\frac{T_f - T_i}{T_i} = \frac{T_f}{T_i}-1##.
 
  • #16
TSny said:
It might be helpful to note that ##\frac{T_f - T_i}{T_i} = \frac{T_f}{T_i}-1##.
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
 
  • #17
Yes, this would give the % change in ω. Likewise, you can determine an expression for the % change in T in terms of just ri and rf.

(Actually, of course, to get % change you need to multiply by 100.)
 
  • #18
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?
 
  • #19
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?
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.
 
  • #20
TSny 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.
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?
 
  • #21
ht9000 said:
ω in terms of T is (Δθ)/T right?
Yes, with Δθ = 2##\pi##.
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?
Yes, that's it.
 
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  • #22
TSny said:
Yes, that's it.
Awesome, thanks a lot for the help
 
  • #23
OK. Good work.
 

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