Recent content by Shuo Xue

I also cannot assume that the final potential energy is 0 so I think this is the most understandable solution. mgh + 1/2mv^2 + 1/2IW^2 = mgh + 1/2mv^2 + 1/2IW^2 mgh + 1/2mv^2 + 1/2mv^2 = mgh + 1/2mv^2 + 1/2mv^2 gh + 3/4v^2 = gh + 3/4v^2 (9.81)(30.58) + (3/4)(20^2) = (9.81)(20.58) + 3/4v^2...

Although I got the correct answer for that, I still don't understand why do I have to add 10m to the height that I got instead of subtracting 10m to it.

The wheel is already rolling on the top of a hill my mistake is I assume the initial kinetic energy is 0 but it is actually not. So initial energy is mgh + 1/2mv^2 + 1/2IW^2mgh + 1/2mv^2 + 1/4mv^2 = 1/2mv^2 + 1/4mv^2 gh + 1/2v^2 + 1/4v^2 = 1/2v^2 + 1/4v^2 gh + 3/4v^2 = 3/4v^2 (9.81)(10) +...

h = 30.58m I tried adding 10m to h instead of subtracting 10m so, h = 40.58 v^2 = 4gh/3 v^2 = 4(9.81)(40.58)/3 v = 23m/s Doing it this way, I got the correct answer.

Homework Statement [/B] A wheel, of radius 200mm, rolls over the top of a hill with a speed of 20m/s and negligible friction losses. (I = 1/2mr^2) Homework Equations [/B] Find the speed of the wheel when it is 10m below the top. The Attempt at a Solution [/B] mgh = 1/2mv^2 + 1/2IW^2 W=...

I've just got into a university and taking Foundation in Chemical Engineering. Have a lot of questions about physics. So I'm here.