Question

A solid 0.595-kg ball rolls without slipping down a track toward a loop-the-loop of radius R = 0.7350 m. What minimum translational speed vmin must the ball have when it is a height H = 1.091 m above the bottom of the loop, in order to complete the loop without falling off the track?

Answer 1

let v be the speed at the top point,, then from gravity relation we have

v^2/r = g

v = sqrt(g*r)

velocity v = sqrt(9.8*0.7350) = 2.68 m/s

let now u is the speed at height H.

Work done by gravity = change in kinetic enrgy

W = 0.5*m*u^2 + 0.5*I*w^2 - (0.5*m*v^2 + 0.5*I*w^2)

W = (0.5*m*u^2 + 0.5*(2/5)*m*r^2*w^2 - (0.5*m*v^2 + 0.5*(2/5)*m*r^2w^2))

W = 0.7*m*u^2 - 0.7*m*v^2 = 0

m*g*(2*R - H) = 0.7*m*(u^2-v^2)

g*(2*R - H) = 0.7*(u^2-v^2)

u^2 - v^2 = g*(2*R - H)/0.7

u = sqrt(v^2 + g*(2*R-H)/0.7)

intial speed u = sqrt( 2.68^2 + 9.8*(2*0.7350 - 1.091)/0.7)

initlal speed u = 3.533 m/s