Recent content by slambert56

Okay so I think the answer should be 180N/m unless the book is wrong and I got 114N/m. Now if there is a mistake in my work shown in the picture it would have to be the height final or height initial. I am going with height inital is the x initial and the height final equals the x final.(I think...

ahhhhh okay when i typed it into my calculator as (2pie) and it didnt multiply apparently. Its 974. Thanks for catching my dumb mistake. I thought it would automatically multiply but I have to put in the sign.

I just divided 6120 by 2pie just to convert radians into revolutions. 1 rev= 2 pie radian

Ok the answer in the back of my book says 974 revolutions and a very short hand way to get that is divide the big wheel radius by the small wheel radius to get 3.53 so i can just multiply 3.53 times the 276 rev to get 974 rev. But I tried a long way using the s=rtheta formula b/c on a test I...

Thanks red and Doc I was making things more complicated then they should have been. It seemed weird at first that the normal force and push force had the same lever arm but I see it now. http://img847.imageshack.us/img847/4014/001hxy.jpg Uploaded with ImageShack.us finally got it.

I think it is one of those frictionless problems. I forgot to post the actual question ha. the drawing shows a bicycle wheel resting against a small step (height=.120m). The weight and radius of the wheel are w=25N and r=.340m. A horizontal force F is applied to the axle of the wheel. As the...

ok this was a problem in my book and the answer apparently is 29N. I have the diagram and pictures below. I know that the pivot point will be the edge of the block that is touching the outside of the wheel. When the wheel is in equilibrium the sum of the torque will=zero so i can set up the...

Thanks for the help. I usually make little mistakes so my plan is to derive the relative velocity because it is faster and applies to all elastic collisions. I can always do both to check my work though.

http://img535.imageshack.us/img535/8316/004plx.jpg Uploaded with ImageShack.us I am going to frame this on my wall haha

yes! now I will surprise my teacher with a curveball. :)

ahh i saw another post from you ok so I can just basically use the KE final=KE initial and solve for vo1-vo2 ?

gneill I feel like I have found the holy grail. That so makes it easier. Thanks. would you happen to know of a website explaining the theory and all. If i use it on a test I will want to explain it.

ok thanks for that error but I still get 5m/s http://img831.imageshack.us/img831/4113/003lzn.jpg Uploaded with ImageShack.us gneill so i could just say that -(v01+v02)=vf1+vf2 ? I think I am a confused on that one. Isnt what you said pretty much the momentum conservation?

Thanks for the help guys. Yes I am given the two masses and 1 ball is moving at 5m/s and the other is at rest. For some reason I keep on getting over 3m/s. I simplified the masses in this one to make it faster because I am all about finding the easiest way...

I did that but I still couldn't get the final velocity of one to be just vf squared. I always seemed to come up with a vf^2 and a vf which led to the quadratic equation again.