How Fast Does the Rim of a Spinning Disk Travel?

[Nitrag]

Problem 13.30
A thin, 60.0g disk with a diameter of 7.00 cm rotates about an axis through its center with 0.240 J of kinetic energy. What is the speed of a point on the rim?

Homework Equations

m=.06kg
r=.035m
KE=.240J

The Attempt at a Solution

K=(1/2)*I*(w^2)
I=(1/2)*M*R^2

I'm using the formula above but I'm not getting the right answer. My numbers seem way to small. I eventually get w(omega) alone. Where do I go from here. Do i need to solve for the length to figure out rad/s then convert to m/s?

Perhaps I am approaching this wrong.

 

[Nitrag]

Calculation error, sorry. Mods you can delete.

 

[ogb p]

After further consideration, I believe the correct approach would be to use the formula for rotational kinetic energy, K=(1/2)*I*(w^2), where I is the moment of inertia and w is the angular velocity. In this case, the moment of inertia for a thin disk rotating about its center is I=(1/2)*m*r^2, where m is the mass of the disk and r is the radius. Plugging in the given values, we get:

K=(1/2)*(0.06kg)*(0.035m)^2*(w^2)=0.240J

Solving for w, we get w=92.1 rad/s. To convert this to linear speed, we can use the formula v=r*w, where v is the linear speed and r is the radius. Plugging in the given values, we get:

v=(0.035m)*(92.1 rad/s)=3.23 m/s

Therefore, the speed of a point on the rim of the disk is 3.23 m/s. It is important to note that this is the speed of a point on the rim, and not the average speed of the entire disk.