Particles of mud are thrown from the rim of a rolling wheel. If the forward speed of the
wheel is v0, and the radius of the wheel is b, show that the greatest height above the ground that the mud can go is
b + v02 / 2g + gb2/ 2v02
At what point on the rolling wheel does this mud leave?
(Note: It is necessary to assume that v02≥bg.)
In 3 dimensions concept,
r = i b cosθ + j b sinθ
Since v = rω = rθ',
v = dr / dt = (-b sin θ i + b cos θ j) θ' = -v0 ( -sin θ i + cos θ j )
The Attempt at a Solution
Firstly, find the time when the particle meets maximum height.
Take downwards as positive
for vertical direction,
v = u + at
0 = v0 cos θ + g t
Then I find the time with minus sign.
It sounds quite weird.
In case, how do I know which side of rim the particle is thrown from rolling wheel?
The forum has already posted that problem already but still I have no idea with it.