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tangibleLime
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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
[tex]C = \frac{1}{2}MR^2[/tex]
[tex]K_{rot}=\frac{1}{2}I\omega^2[/tex]
The Attempt at a Solution
Since the formula for kinetic rotational energy is [tex]K_{rot}=\frac{1}{2}I\omega^2[/tex], and the constant for the moment of inertia for a solid disk with the axis of rotation about it's diameter is [tex]C = \frac{1}{2}MR^2[/tex], I substituted the I in the second equation with the first equation, resulting in the following:
[tex]K_{rot}=\frac{1}{2}(\frac{1}{2}MR^2)\omega^2[/tex]
Substituting the values that I was supplied with in the problem statement, I came up with this equation:
[tex]0.190=\frac{1}{2}(\frac{1}{2}(0.05)(0.04)^2)\omega^2[/tex]
Solving for [tex]\omega[/tex], I ended up with [tex]\omega \approx \pm 97.5[/tex], which was determined to be incorrect.
Any help would be extremely appreciated.