Question

Suppose a solid sphere of mass 450 g and radius 5.00 cm rolls without slipping down an inclined plane starting from rest. The inclined plane is 7.00 m long and makes an angel of 20.0 o from the horizontal. The linear velocity of the sphere at the bottom of the incline is _______ m/s. please show work

Answer 1

here,

mass , m = 450 g = 0.45 kg

radius,r = 5 cm = 0.05 m

length , l = 7 m

theta = 20 degree

let the linear velocity at the bottom of the incline be v

using conservation of mechanical energy

initial potential energy = final kinetic energy

m * g* l * sin(theta) = 0.5 * m * v^2 + 0.5 * I * w^2

m * g* l * sin(theta) = 0.5 * m * v^2 + 0.5 * (0.4 * m * r^2) * (v/r)^2

g* l * sin(theta) = 0.7 * v^2

9.81 * 7 * sin(20) = 0.7 * v^2

solving for v

v = 5.79 m/s

the linear velocity of the sphere is 5.79 m/s