Recent content by Alan6585

ah Well again, I solved it. I think I understand that I should only use inertia to solve for alpha since that is the net torque and I was leaving out the block or something. Anyway my answer was wrong cause I solved it the wrong way. 3601 was the answer I got, when I got it CORRECT the...

You were right, thanks, even though my book clearly stated that counter-clockwise motion is negative and also it does not ask for magnitude. Thanks.

it gives me gravity in this problem and in part 2 and 3 the system swings along the pivot point due to gravity so I'm sure it comes into play. Also talking about how alpha is being related to the end etc etc.. isn't that the point of solving for the moment of inertia? I may try your method...

alpha * r = acceleration I'm not sure exactly what r should be in this case. I'm pretty sure it would be 35.. but I tried 35/2 also.. it didn't work. maybe you are right actually that's for the tangential component.. the other component being V^2 / r or w^2 r. I'm not sure how this...

shouldn't this be negative since it is clockwise. If so I already tried that answer. Didn't try it postive.

The question shows a horizontal uniform rod with a pivot point on the left end and a weight tied to the right end. The rod and weight have mass 7 kg each. The rod is 35m long It asks what the inital torque of the system on the pivot point will be when dropped with gravity being 9.8 m/s^2...