- #36
postfan
- 259
- 0
Is that right ?
It looks right. Then solve the second equation for the angular acceleration and multiply both sides by R and then plug that expression in for a (because Rα=a like we said) then you can solve for Ff and then you can put Ff back into the original linear acceleration equation to get the answer.postfan said:Ok so 2*M*g*sin(alpha)-F_f=2M*a for forces
and F_f*R=(2/3)*M*R^2*angularacceleration for torque
How's that?
Yespostfan said:So I did that and got a=3/4gsin(alpha). Is that right?
Well, I can't get inside the mind of the test-makers, but if you're asking if that piece of information was irrelevant, then yes, it was irrelevant.postfan said:Also did they just give you the moment of inertia for a solid sphere just to trick you?
Had they given you all the formulae except that one it would have been rather a strong clue.postfan said:did they just give you the moment of inertia for a solid sphere just to trick you?