Recent content by mc8569

I just need an explanation: A spinning bicycle wheel is mounted on an axle you can hold in your hands. You hold it in front of you with the axle horizontal. The direction of the angular momentum vector is to your right. You then rotate the axle clockwise in a vertical plane as seen by you...

Omg.. no one is really giving me much help. The case for when the bullet leaves the block at v/3 is when the block is not fixed, and moves due to the frictional force of the bullet. In c) the block is stated to be fixed, and must use v/2 from part b). The main problem I have is, can we say that...

Can someone pleaseeee help me! I need help by tomorrow... which is when this problem is due! I would appreciate it a lot.. =(

someone pleaseeeeeeeeeeeee help me!

Er, my part b) I am 99.99% sure I am correct. Initial velocity (initial kinetic energy) before entering/collision with the block (before it loses any kinetic energy) is v. Its final velocity when it exits the block (v/2) consists of its final kinetic energy. Therefore, KEi - Wnc = KEf *You can...

glueball, i don't understand what you're talking about. could you please explain to me? I'm very confused..

So are you saying I should use what I found in b), divide it by L thereby solving for the frictional force f. Using this f use work = fdcos@ = KEi and solve for this new d accordingly? Therefore, is this right: From b) solve for frictional force by dividing that found value by L: *can't use a)...

Homework Statement http://i48.tinypic.com/2cfb3py.png ^^Diagram of situation There is a projectile of mass m and initial velocity v fired into a block of width L and mass M at rest on a frictionless surface. a) Find the speed of block M as the bullet emerges from the back side with a...

AHHHH OHHH NEVERMIND! Sorry... *CONICAL* pendulum, lol I was thinking of a regular, 2-D pendulum. I would delete this if I could, but I can't =\ But no need to respond anyone! So silly!

Consider a conical pendulum with a weight on it and that makes some angle with the vertical. I am told to find the radial acceleration of the weight and someone showed me how it is solved but I don't understand why: I am told to set Tsin@ = m(ar) *ar = radial acceleration ar = (Tsin@)/m...

If I integrate it should I be integrating it from Vo to Vo/3? Or do I do something else with the integrated equation for t. I think I will integrate from Vo to Vo/3 that seems logical. Also, could you give me a brief explanation as to why doing: fk = -(umv^2)/R v^2 = -fR/um 2v(dv/dt) =...

should I bring up dt and bring over v^2 to the left and all its coefficients, and integrate that from Vo to Vo/3? like such: dv/dt = -uv^2/R (-R/u)(1/v^2)dv = dt integrate that equation?? It doesn't seem to make much sense though. I'm so clueless..

Doc I've given it thought but I really have no clue what to do, what direction to go in. Should I integrate the equation? I don't know what would that even do though.

Oh.. okay awesome! But I don't get what you mean by the hint about dv/dt will come into play for part c) because how I am solving for c) is quite simple: with a = -(uv^2)/R Vf = Vi + at (1/3)Vo = Vo + (-uv^2/R)t I assume that Vo = v and therefore t = (2R)/(3uv) Is that right...

Yes! Thank you for answering =DDD Now, I did what you said, which is also what I originally asked if I should do and I have dv/dt = -(R/(2umv))(d(fk)/dt) The issue I have, which I presented when I originally asked if I should do that, is with the fact that the answer must be in terms of m, R...