A uniform, solid disk with mass m and radius R is pivoted about a horizontal axis through its center. A small object of the same mass is glued to the rim of the disk.
If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.
Express your answer in terms of the variables m, R and appropriate constants.
2007-12-02 04:59:19 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
I have tried the answers:
square root(2g/R)
and square root (4g/R) and they dont' come out as correct.
2007-12-02 05:20:31 · update #1
Jgoulden, I tried this method, but the answer is coming out wrong for some reason. Are you sure that is correct?
2007-12-02 05:25:29 · update #2
Use energy to solve this one.
The initial energy is entirely potential: the small object with mass m is at height R, so
E = m g R
The final energy is entirely kinetic
E = 1/2 I w^2
where I is the moment of inertia of the system. The moment of inertia of a solid disk about the center is 1/2 m R^2, and the moment of inertia of a mass m a distance R from the center of rotation is m R^2. Plug 'em in, set the two energies equal, and solve for the angular velocity w.
2007-12-02 05:17:44 · answer #1 · answered by jgoulden 7 · 3⤊ 1⤋
dude, use that theoreom where you add like mR^2
easy BS.
2007-12-02 05:13:11 · answer #2 · answered by Anonymous · 0⤊ 1⤋
lol
2007-12-02 05:01:47 · answer #3 · answered by Anonymous · 0⤊ 1⤋