A disc rolls up an inclined plane of angle 45°. The centre of mass of the sphere has a translational speed of 2 m/s at the bottom of the plane. What is the distance through which the disc travel up the plane ?

Explanation

`(1//2)mv^(2) + (1//2)Iomega^(2)=mg s sin 45, (3//4)mv^(2) = mgs//sqrt(2), s = 0.433 m` (distance)