Question

A solid sphere of radius r0=1.5cm rolls without slipping inside a circular track of radius R0=26cm starting from rest at a height R0 above the bottom of the track. When it leaves the track passing through an angle of 1350 as shown in the figure below, (a) what will be its speed when it is launching from the track, and (b) at what distance D from the base of the track will the sphere hit the ground?

135 Ro

Answer 1

here,

r0 = 0.015 m

R0 = 0.26 m

theta = 135 degree

a)

let the speed of launching from the track be v0

using conservation of energy

0.5 * m * v0^2 + 0.5 * I * w^2 = m*g * ( r0 * sin(45))

0.5 * m * v0^2 + 0.5 * 0.4 * m * r0^2 * (v0/r0)^2 = m*g * ( r0 * sin(45))

0.5 * v0^2 + 0.2 * (v0)^2 = g * ( r0 * sin(45))

0.7 * v0^2 = 9.8 * 0.26 * sin(45)

v0 = 1.6 m/s

the launching speed is 1.6 m/s

b)

the angle of launch , phi = 90/4 = 22.5 degree

the distance D from the base be D

D - r0/2 = v0^2 * sin(2 * theta) /g

D - 0.26/2 = 1.6^2 * sin(45) /9.8

D = 0.31 m