any physicists around? need help on a prob

[churchdoesmatter]

A uniform solid disk with mass m and radius R is pivoted about a horizontal axis through its center, and a small object of the same mass m is attached 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 velocity when the small object is at the bottom. (Use g, m and R as necessary.)

any help will be appreciated. thanks.

 

[silverpig]

Your rotational energy (1/2*I*omega^2) of the disk + rotational energy of the small mass (same formula, but instead of finding I you can just simplify to straight linear motion... 1/2 * m * v^2) = gravitational energy difference of small object over a distance of 2*R

 

[silverpig]

Eh, forget that simplification I said about using v... you want omega, so you have to use it in that part too... realize that omega for the mass is the same as for the disk as they are attached.

 

[churchdoesmatter]

so it would be (1/2)I*w^2 + (1/2)mv^2 = ??? .. can you write it all out for me please? im confused

 

[churchdoesmatter]

anyone else can take a shot at it too.. any help will be appreciated. thanks.

 

[Akira13]

Assuming no friction (or other non conserved forces), isn't the potential energy turned into kinetic energy? It's been almost two years since my last physics class, but I seem to remember mgh=PE, where h is 2R (if I understand the problem correctly). You can use that PE value to calculate the kinetic energy (thus, velocity) of the object.

 

[churchdoesmatter]

the answer i get is sqrt(8g/3R) . which is wrong . can someone else give me some input? thx

 

[WarmAndSCSI]

quite an interesting topic coming from "pokemonlover"

LOL at the 9-year-old

 

[Akira13]

Originally posted by: pokemonlover
the answer i get is sqrt(8g/3R) . which is wrong . can someone else give me some input? thx

So what's the right answer? Should it involve pi (since we're talking about angular velocity: radians/sec)?