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 ******************************** * * * horizontal distance = 6.4m * * * ********************************