What Is the Correct Angular Speed for Doubling the CD's Kinetic Energy?

[UCrazyBeautifulU]

A 10.7 g CD with a radius of 6.06 cm rotates with an angular speed of 30.9 rad/s. What is its kinetic energy?

What angular speed must the CD have if its kinetic energy is to be doubled? Here is my work:

kinetic energy energy = 1/2 m v2
= 1/2 m ( r ω )2
= 0.5 x 0.0107 x ( 0.0606 x 30.9 )^2 J

angular speed must be increased by ( 2)^ 1/2 if kinetic energy is duobled
since kinetice energy is proportional to ω2

That answer isn't right, it comes out to be 0.188 J, can anyone tell me what is wrong with my equation?

 

[OlderDan]

A 10.7 g CD with a radius of 6.06 cm rotates with an angular speed of 30.9 rad/s. What is its kinetic energy?

What angular speed must the CD have if its kinetic energy is to be doubled?


Here is my work:

kinetic energy energy = 1/2 m v2
= 1/2 m ( r ω )2
= 0.5 x 0.0107 x ( 0.0606 x 30.9 )^2 J

angular speed must be increased by ( 2)^ 1/2 if kinetic energy is duobled
since kinetice energy is proportional to ω2

That answer isn't right, it comes out to be 0.188 J, can anyone tell me what is wrong with my equation?

Chek your formula for the kinetic energy of a rotating rigid body. Not everything is moving with the speed you used.

 

[UCrazyBeautifulU]

what is wrong with the formula? Only one speed is given.

 

[Astronuc]

See this for discussion of rotational kinetic energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html

One angular speed is given, but the local linear speed depends on r to which OlderDan alluded, i.e. v = $\omega$r.

One must use the appropriate moment of inertia for the CD.

 

[UCrazyBeautifulU]

Thanks for all the wonderful help!