Rotating Disk

  • #1

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.
 

Answers and Replies

  • #2
258
1
It asks for the SPEED of a point on the rim. It does not ask for the angular speed.
 
  • #3
I figured it was something like that, so shortly after posting the initial topic, I tried [tex]v=r\omega[/tex] as [tex]v=(0.04)(97.5)[/tex] which results in the correct answer of 3.9. Originally I recieved 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!
 

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