How high is the upper end of the inclined plane?

A disk released from the top of an inclined plane has a speed of 7.12 m/s at the bottom. How high is the upper end of the inclined plane? Assume that the disk rolled without slipping.How high is the upper end of the inclined plane?NOTE: m = mass, g = gravity, h = height, v = velocity, I = moment of inertia, 蠅 = angular velocity, r = radius



I'm surprised no one has answered this question yet. Anyway, conservation of energy tells us energy in a closed system remains constant. In this case, original potential energy = final kinetic energy (or PE = KE)



For rolling motion, KE is (1/2)m*v虏 + (1/2)I*蠅虏. PE is still m*g*h.

PE = KE

mgh = (1/2)m*v虏 + (1/2)I*蠅虏



Angular velocity (蠅) is v/r. Moment of inertia (I) of a disk is (1/2)m*r虏. Plug those in and simplify.

mgh = (1/2)m*v虏 + (1/2)I*蠅虏

mgh = (1/2)m*v虏 + (1/2)(1/2)m*r虏(v/r)虏

mgh = (1/2)m*v虏 + (1/4)mv虏

gh = (1/2)v虏 + (1/4)v虏

gh = (3/4)v虏



Plug in the values and solve;

gh = (3/4)v虏

(9.8)*h = (3/4)(7.12)虏

h = 3.88 meters