Question

The new angular velocity is

Answer 1

a. 41
b. 4.0
c. 39
d. 36
e. 6.0

Answer 2

Ok so correct me here...

I'm first trying to look for KE by filling in
.5*8g*(6*4)^2 = KE

For velocity should I multiply 6m*4rads/sec?

Answer 3

Then, do I use this to find omega?
\[2304=\frac{ 1 }{ 2 }*8g*2m^2*\omega^2\]

Answer 4

Forgive me. I had my head up my retriceas I was solving multiple other problems at the same time.

Let's conserve angular momentum. Angular momentum is:\[L=I \omega \]where L is angular momentum; I is moment of inertia; and ω is angular velocity. Now insert the value for the angular momentum of a point mass at distance r from the axis of rotation and we get:\[L=mR ^{2}\omega \]Now there are two different states: one with R1=6m and ω1=4 rad/s; and one with R2=2m and ω2 unknown. Since angular momentum is conserved we have:\[L _{1}=L _{2}\]\[mR _{1}^{2}\omega _{1}=mR _{2}^{2}\omega _{2}\]Now solve for ω2 and we get:\[\omega _{2}=\frac{ mR _{1}^{2} \omega _{1}}{ mR _{2}^{2} }\]Reduce that and you get:\[\omega _{2}=(\frac{ R _{1} }{ R _{2}})^{2}\omega _{1}\]

I always find that it's easier to simplify the equations as much as possible before inserting numbers.

Answer 5

Kinetic energy is not conserved in this question; this is why you have to work with angular momentum as PsiSquared showed you.

Answer 6

If you're wondering why energy isn't conserved in this problem, why the energy actually increases when the mass's radius of revolution is shortened, it's because you have to do work on the system--i.e. add energy to the system--to shorten the radius of revolution. Remember there is a centripetal force created in the string by the rotating mass. When you shorten that string you have to pull on it with some force and then pull it for some distance. Any force applied over a distance is work, and in this case that work adds to the energy of the system.

Answer 7

Thanks!