Recent content by mdk31

No, that looks correct.

I see, that makes sense. Thanks!

I just looked at this problem again and realized that it also wants us to find the speed of the bullet just before impact. The angular speed of the system immediately after impact is given as 4.5 rad/s. I know I'm supposed to use conservation of angular momentum rm[0]v[0]sin(x)=Iw (w is...

How do you do them? Very carefully. You can read this and see if you get some insight. http://www.sparknotes.com/physics/linearmomentum/collisions/section2.rhtml

Why use integration by parts? Why not just use substiution? u=x+2 du/dx=1

Re-read your question. The pivot point is no longer at the point where the weight force is applied; it is at the right side where the right tension is applied. If it was still at the point where the weight is applied, you would have the same answer from your first question as before and you...

The distance from where the weight force is applied to the pivot point is (L-x) but I do not see that in your equation. Also, the tension in the left side is not (L-x) from the pivot point and the sign on the weight is wrong. I know it must be mind-bogglingly frustrating that I am not...

No, because the tension from the right side does not contribute to the torque if it is applied at the pivot point. And also, what is the distance from the place where the weight is applied to the pivot point?

Imagine someone pulling up on the left side of the rod; the other side would go down right? It would rotate clockwise. Now consider if someone was pulling up on the right side of the rord. The other side would go down and the rod would be rotating counterclockwise. It is the same concept...

Why are you dividing by L?

Think about what forces in the problem will contribute to net torque. If you choose the pivot point to be at the the place where the string is attached, you know that it will not count towards net torque. What about the center of mass of the rod? Does the gravitational force on it not count...

Not necessarily. For example, if an object moves from position 0 m to position 300 m, and from position 300 m to a final position -290 m, the total displacement is -290 but the total distance traveled is certainly not +290. The total distance traveled would be 890

These numbers tell you that you need a greater force to get the car sliding than you need to keep it sliding.

Do you get the same answer when you try conservation of energy?

The force of gravity does not contribute to the acceleration.