Question

A solid sphere of mass 0.6010 kg rolls without slipping along a horizontal surface with a translational speed of 5.420 m/s. It comes to an incline that makes an angle of 31.00° with the horizontal surface. To what vertical height above the horizontal surface does the sphere rise on the incline?

Answer 1

let m is the mass and r is the radius of the sphere.

let v = 5.420 m/s

let w is the angular speed of the sphere.

Apply conservation of energy

initial linear and rotaionalkinetic energy = final potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*H

(1/2)*m*v^2 + (1/2)*(2/5)*m*r^2*w^2 = m*g*H

(1/2)*m*v^2 + (1/5)*m*(r*w)^2 = m*g*H

(1/2)*m*v^2 + (1/5)*m*v^2 = m*g*H (since v = r*w)

(7/10)*m*v^2 = m*g*H

==> H = 7*v^2/(10*g)

= 7*5.42^2/(10*9.8)

= 2.0983 m <<<<<<<<<<----------------Answer