CATAGORY: Mechanics
QUESTION: A wheel is being rolled along a level path with a constant
speed of ten m/s. A small lump of mud stuck on the wheel suddenly breaks
losse when it is at the top of the wheel. If the wheel has a diameter of
two m, how far ahead of the axle is the lump when it strikes the path?

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At the top of the wheel the velocity of a point relative to the ground is
twice the velocity of the axle.  The x component of the mud lump V(x,mud)
is

(1)      V(x,mud) = 2 V(x,axle) = 2 X 10m/s =20m/s

Once free of the wheel the mud lump falls as if dropped. the y position of
the mud lump is given by

(2)       y = y(initial) - 0.5gt^2

y(initial) = 2 meters 
g = 9.8m/s^2

The final value for y is y = 0m. (ie when the mud lump strikes the ground)
thus    

(3)      0m = 2m -0.5gt^2.

Solving for t gives

(4)        t = sqrt (2m / 0.5g) = sqrt(2m / 0.5X9.8m/s^2) = .64s

           the distance the mud lump travels horizontaly is then
           
(5)           x(mud) = V(x,mud) X .64s = 20m/s X .64s = 12.8m

But during this time the axle advances a distance

(6)           x(axel) = V(x,axle) X .64s = 10m/s X .64s = 6.4m

the horizontal distance between the lump and the axle when the lump
strikes the ground is

(7)           horizontal distance = x(mud) - x(axle) = 12.8m-6.4m = 6.4m

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              *  horizontal distance = 6.4m  *
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