Question

I need to find velocity of center of mass of cylindricall log rolling downhill.

this is what i have

K=1/2mv^2+(1/2)I?^2

=1/2mv^2+(1/2)I(v/r)^2

=1/2(m+I/r2)v^2

mgh=(1/2)(m+((1/2)mr^2)/r^2)v^2

I dont know if i am heading in the right direction or how to simplify

Answer 1

Now consider an object rolling down an incline plane. The first thing that we are going to do is draw a freebody diagram of the forces acting on this mass and then resolve those forces into their components which act parallel and perpendicular to the plane.

Let's consider the axis of rotation passing through the disk's center of mass, cm. Notice in this case that only the instantaneous static friction force will supply a torque since the lines of action of the other two forces (normal and weight) act through the center of mass and cannot produce a torque. As long as the mass rolls without slipping, we can use the relationships: v = r? and a = r?.

Rotationally,

net ? = I_CM?
? = f_sr

f_s = I_CM(?/r)

Remember linearly,

net F = ma
mg sin? - f_s = mr? where a = r?

Simultaneously eliminating f_s and solving for ? yields:

mg sin? - I_CM(?/r) = mr?
? = g sin? /(r + I_CM/mr)

The moment of inertia for a disk, or solid cylinder (see chart provided below), equals