# Fnet=ma 2008 #14

1. Jan 28, 2009

### dtl42

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Not sure.

3. 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.

2. Jan 28, 2009

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

3. Jan 28, 2009

### dtl42

Thanks, I get it now.

4. Jan 29, 2009

### compwiz3000

Can somebody explain this in detail?

5. Jan 30, 2009

### 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 $r_{2}$,

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