Uniform Sphere Rotation: Kinetic Energy and Torque

[veronicak5678]

Homework Statement

The Earth can be approximated as a solid, uniform sphere with r = 6380 km.
a- Using 1 day = 24 hrs, find the rotational kinetic energy.
b- It revolves around the sun at a distance of 1.5 x 10 ^ 11 m once every 365.25 days. Find kinetic relative to tsun.
c- If we wanted to increase the day to 25 hours, what torque would be required in 1 year to do this?

Homework Equations

The Attempt at a Solution

I am using mass = 5.97 x 10^24 kg

a- 2.58 x10^29 J
b- 2.66 x 10 ^33 J
c- ?

 

[tiny-tim]

Hi veronicak5678!

(have an omega: ω )

The Earth can be approximated as a solid, uniform sphere with r = 6380 km.
a- Using 1 day = 24 hrs, find the rotational kinetic energy.

What is the formula for rotational KE?

What is the moment of inertia of a sphere?

 

[veronicak5678]

RK = 1/2 I w^2

I = 2/5 MR^2

(w for omega, angular velocity)

 

[tiny-tim]

RK = 1/2 I w^2

I = 2/5 MR^2

(w for omega, angular velocity)

Hi veronicak5678!

(what happened to that ω I gave you?)

ok, now torque = rate of change of angular momentum …

so what is the change in angular momentum if the day is lengthened to 25 hours?

 

[veronicak5678]

Sorry! I didn't even see that little alphabet.
So the desired ω is .0000698 rad/s to make a 25 hour day. I guess by change in momentum you mean the same as change in velocity? .0000727 - .0000698 rad/s ?

 

[tiny-tim]

So the desired ω is .0000698 rad/s to make a 25 hour day. I guess by change in momentum you mean the same as change in velocity? .0000727 - .0000698 rad/s ?

No, angular momentum is Iω.

I mean the change in Iω.

 

[veronicak5678]

OK. Actual Iω is 9.77 x 10^37 kg m ^2 * .0000727 rad /s = 7.11 x 10^33.
Desired Iω is 9.77 x 10^37 kg m ^2 * .0000698 rad /s = 6.82 x 10^33


Difference of 2.90 x 10^32

Not sure if that is what you meant...