What is the speed of a disk rolling down an inclined plane?

Question

A solid disk of radius 5 inches and weighing 5kg is released from rest and rolls without slipping down an inclined plane 25 degrees to the horizontal. What is the speed of the disk after rolling 5metres down the plane?

Answer 1

♣ its pot energy was E=mgh, where h=(5 m)*(sin 25°);

♦ its kin energy became E=E1+E2, where E1=0.5*m*v^2 is kin energy of translation, E2=0.5*J*w^2 is kin energy of rotation, J=0.5m*r^2 is its moment of inertia, v=w*r is linear speed, w is angular speed of the disk, r= 5in = 5*25.4e-3 m, m=5kg, g=9.8 m/s^2;

♥ E=E; mgh =0.5*m*v^2 +0.5*J*w^2;

mgh =0.5*m*v^2 +0.5*0.5m*r^2*(v/r)^2;

2gh = v^2 +0.5*v^2;

v=√((4/3)*g*h) = 2√((9.8 *5 *sin25)/3) =5.25 m/s;

as you can see neither m nor r are necessary!