Question

A wheel of radius b is rolling along a muddy road with a speed v. Particles of mud attached to the wheel are being continuously thrown off from all points of the wheel. If v2 > 2bg, where g is the acceleration of gravity, find the maximum height above the road attained by the mud, H = H(b,v,g).

Answer 1

Concepts: Energy conservation, projectile motion,
	Reasoning: The mud particles are projectiles acted on by the gravitational force. Mud particles thrown from different positions on the rim follow different trajectories and rise to different heights. We will label the position on the rim by some parameter ?, calculate the height reached by a particle leaving from this position, and then find the maximum height as a function of the parameter ?.
	Details of the calculation: Assume mud is ejected from a point P, and define ? and h as in the figure. Let ?h be the height the mud rises above the ejection point. Conservation of energy yields mg?h = (mvy2)/2 = (mv2 cos2?)/2, ?h = (v2 cos2?)/2g. The height H reached by the mud is H = h + ?h = b + b sin? + (v2 cos2?)/2g. We find the maximum height by setting dH/d? = 0. This yields b cos? - (v2/g) cos? sin? = 0. Two solutions to this equation exist. (i) cos? = 0, ? = ?/2, H = 2b. This is the maximum height only if b sin? + (v2 cos2?)/2g ? b for all angles ?, or v2 ? 2gb. Then we do not have v2 > 2gb. (ii) sin? = gb/v2 = sin?0. Then H = b + b sin?0 + v2(1 - sin2?0)/(2g) H = b + gb2/v2 + (v2/2g)(1 - g2b2/v4)