Speed of Point on Rim of 50.0 g Disk with 8.00 cm Diameter

[tangibleLime]

Homework Statement

A thin, 50.0 g disk with a diameter of 8.00 cm rotates about an axis through its center with 0.190 J of kinetic energy. What is the speed of a point on the rim?

Homework Equations

$$C = \frac{1}{2}MR^2$$
$$K_{rot}=\frac{1}{2}I\omega^2$$

The Attempt at a Solution

Since the formula for kinetic rotational energy is $$K_{rot}=\frac{1}{2}I\omega^2$$, and the constant for the moment of inertia for a solid disk with the axis of rotation about it's diameter is $$C = \frac{1}{2}MR^2$$, I substituted the I in the second equation with the first equation, resulting in the following:

$$K_{rot}=\frac{1}{2}(\frac{1}{2}MR^2)\omega^2$$

Substituting the values that I was supplied with in the problem statement, I came up with this equation:

$$0.190=\frac{1}{2}(\frac{1}{2}(0.05)(0.04)^2)\omega^2$$

Solving for $$\omega$$, I ended up with $$\omega \approx \pm 97.5$$, which was determined to be incorrect.

Any help would be extremely appreciated.

 

[zachzach]

It asks for the SPEED of a point on the rim. It does not ask for the angular speed.

 

[tangibleLime]

I figured it was something like that, so shortly after posting the initial topic, I tried $$v=r\omega$$ as $$v=(0.04)(97.5)$$ which results in the correct answer of 3.9. Originally I received an answer that was something like 360 (I must have entered a wrong number or something into the calculator) so I ignored it. Then after you posted and backed up my thoughts, I tried again and got the correct answer. Thanks!