Fnet=ma 2008 #14: Kinetic Energy Increase with Angular Velocity

[dtl42]

Homework Statement

A spaceborne energy storage device consists of two equal masses connected by a tether and rotating about their center of mass. Additional energy is stored by reeling in the tether; no external forces are applied. Initially the device has kinetic energy E and rotates at angular velocity ω. Energy is added until the device rotates at angular velocity 2ω. What is the new kinetic energy of the device?

Homework Equations

Not sure.

The Attempt at a Solution

I guessed that since kinetic energy is related to angular velocity squared, it would be 4E, but the answer is 2E.

 

[Dadface]

v=rw and if the tether is reeled in r reduces as w increases.Angular momentum is conserved.

 

[dtl42]

Thanks, I get it now.

 

[compwiz3000]

Can somebody explain this in detail?

 

[bowma166]

Well, as you're pulling in the tether, angular momentum is conserved. Angular momentum: $L=mvr$, so:

$$mvr_{1}=mvr_{2}$$
rewriting v as wr...
$$\omega_{1} r_{1}^{2}=\omega_{2} r_{2}^{2}$$

With $\omega_{2}=2\omega_{1}$, we get

$$r_{1}^{2}\omega_{1}=r_{2}^{2}\left(2 \omega_{1}\right)$$

$$\frac{r_{1}}{\sqrt{2}}=r_{2}$$


Kinetic energy...
$$K_{1}=mv^{2}=m\left(\omega r\right)^{2}=m\omega^{2}r^{2}$$

So, replacing with $\omega_{2}$ and $r_{2}$,

$$K_{2}=m\left(2\omega\right)^{2}\left(\frac{r}{\sqrt{2}}\right)^{2}=m\left(4\omega^{2}\right)\left(\frac{r^{2}}{2}\right)=2m\omega^{2}r^{2}=2K_{1}$$