Recent content by Speedking96

When x = 0, the value would be -a, and when x = a, the value would be 0. Wow, it was that simple! Hehe.

Like, when the bounds are numbers, I have no problem changing the bounds when doing the substitution, but here I can't seem to wrap my head around it since the bounds are not 'numbers'.

Okay, when you say: After making the substitution u = x - a, the bounds of the u-integral are obtained by substituting the bounds of the x-integral (0 ≤ x ≤ a) into the expression for u. How would you substitute an inequality (0 ≤ x ≤ a) into an equation (u = x - a) ? Thanks for the help.

Homework Statement This is the problem with the solution: Can someone please explain how the new bounds were computed, I don't quite understand what's going on with the inequalities? Also, in the final two steps, how can the f(u) change to f(x)?

Homework Statement You are standing on your skateboard, which is at rest. Your friend throws a heavy ball at you. What should you do to minimize your velocity? A. Catch the ball B. Hit the ball back with the same velocity C. Neither, they both have the same effect. I would say (A). Because...

I actually don't know what you're referring to as I am learning torque and rotational dynamics on my own (on my final exam tomorrow and my teacher rushed everything on rotation ) - I don't think I've come across a problem like that

Wow. The pivot point is the location of the foot and N1 is for the hands. I see. Thank you.

Homework Statement I have a question about the following question: Firstly, shouldn't there be a normal force for each hand (part a), but here they treat it as one normal force. Also in part b, why isn't the torque for Fn1h = 0? Since that is the pivot point, isn't the moment arm zero?

Alright, I understand. Thank you.

Homework Statement Below is the question: I only have an issue with the last step of the problem. Why wouldn't you factor in the translational AND rotational energy of the ball and then solve for maximum height?

Thank you.

The angular momentum for a ring is MR2. Since a ring is a rigid body, it's angular momentum is I = lw = (MR2)(w) = (MR2)(v/r) = mvr Ok, I see. So from what I can understand from the posts above, you can use r x p for a particle, or for a sum of particles, but for a rigid body, you must use I...

Homework Statement I was reading the textbook section on angular momentum, and I'm having some difficulty grasping angular momentum. Here is a question: In the book, it says that the angular momentum L is equal to vector r cross vector p for a particle. But, for a rigid body, the equation...

Ok, I understand. Thank you very much.

Ok. I think I get it; the cosine is used to get the moment arm in this case, correct?