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
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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
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Find the speed of the wheel when it is 10m below the top.
The Attempt at a Solution
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mgh = 1/2mv^2 + 1/2IW^2
W=...