Recent content by Bonbon32

Makes sense so (meu mg/R^2 )* r^2 *Dr which gives me a torque, not sure where they got the 2 from? Which is on the numerator next to the meu value. Finally got one more question if you can bear with me. Would the meu mg the friction be dF instead as we are taking only a small circular position...

Not sure how they obtained an answer of (2meukenetic *Mg/R^2)*r^2 dr = dt When they say disk I am assuming it's not ring shaped. So since F * Dr= DT where I guess F acting on a small point on the disc is constant from the table top. And Dr is the variable distance from a random small point from...