Posted by Hil on Friday, December 6, 2013 at 5:54pm.
This makes no sense.
remember that the sum of the torques is equal to I*alpha
so sigma(t) = r1xF1 + r2xF2 +... +rnxFn = I*alpha
where I is the moment of inertia through the axis of rotation. look for I in a table, easily found on google. notice the similarity to the application of Newtons second law.
set up your problem like that and then just solve for alpha.
remember to set signs for positive and negative torques, i like sticking to a positive CCW, so that when all the analysis is done, the direction of rotation resolves with the sign automatically.
since the moment of inertia depends on the axis of rotation, part b requires the parallel axis theorem to get the moment of inertia through the new axis of rotation, although some tables include the moment of inertia you're looking for already calculated for you.
watch out for the fact that if a force's line of action intersects with the axis of rotation, the force will not cause a torque because r=0 so rxF=0. Remember this paragraph while setting up part B
once you've found your new moment of inertia. the same idea that worked for part A will help you solve for alpha in part B.
part C is a bit more laborious but if you solved A and B correctly, it's just more of the same. as opposed to part B, all three forces provide a torque.
all three forces are applied tangentially to their Rs in all three cases so theres no need for the cross product in any part of this problem.
so, algorithmically:
1.- find out the pertinent moment of inertia
2.- set up your problem as following:
sigma(t)= I*alpha = (r1 x F1) + (r2 x F2) +... + (rn x Fn)
3.- get your alpha algebraically.
if the algebra messes you up, google wolfram alpha
Good luck
a) alpha= 2F/mr ccw
b) alpha= 2F/mr cw
c) alpha= 8F/3mr ccw