Question

A solid sphere starts from rest at the upper end of the track, as shown in figure below, and rolls without slipping until it rolls off the right-hand end. If H = 10.0 m and h = 5.00 m and the track is horizontal at the right-hand end, how far horizontally from point A does the sphere land on the floor?

Answer #1

from the conservation of energy

potential energy = linear kinetic enerrgy+rotational kinetic energy

mg(H-h) = 1/2mv^2+1/2Iw^2

for solid sphere, moment of inertia I = 2/5mr^2

angular speed ,w = v/r

mg(H-h) = 0.5mv^2+0.5*2/5*mr^2w^2

mg(H-h) = 0.5mv^2+0.2mv^2 = 0.7mv^2

v = sqrt(g(H-h)/0.7)

v = sqrt(9.8(10-5)/0.7)

v = 8.37 m/s

time taken to reach the ground from the point A

h = 1/2gt^2

t = sqrt(2h/g)

t = sqrt(2*5/9.8) = 1.01 sec

distance = x = v*t = 8.37*1.01 = 8.45 m

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