Calculate the speed of the DISK at the BOTTOM of the inclined plane?

GPE is converted to KE in this process.

Two modes of storage of kinetic energy exist: translational and rotational.

KEtrans = 1/2*m*v^2

KErot = 1/2*I*omega^2

no-slip condition:

v = omega/r

Thus:

KErot = 1/2*I*v^2/r^2

Thus:

KEnet = KEtrans + KErot

KEnet = 1/2*m*(1 + I/(m*r^2))*v^2

By conservation of energy:

GPEinit = KEnet

Thus:

m*g*h = 1/2*m*(1 + I/(m*r^2))*v^2

Cancel m:

g*h = 1/2*(1 + I/(m*r^2))*v^2

Solve for v:

v = sqrt(2*g*h/(I/(m*r^2) + 1))

For a hoop, I = m*r^2

For a disc, I = 1/2*r^2

Substitute and simplify:

v_hoop = sqrt(g*h)

v_disc = sqrt(2*g*h/3)

Data:

h:=0.66 m; g:=9.8 N/kg;

Results:

v_hoop = 2.543 meters/second

v_disc = 2.077 meters/second

You can find the potential energy of the hoop as m*g*h which would be the magnitude of the mechanical energy. Therefore if you add the kinetic energy and the rotational kinetic energy at the bottom of the ramp they should equal the potential energy at the top. You then get this equation:

mgh = 0.5mv^2 + 0.5Iw^2 where I = moment of Inertia ; w = angular velocity

then by plugging in I = mr^2 ; w = v/r and thusly cancelling out the masses we get:

gh = 0.5(v^2 +rv)

then we put this in quadratic form to get:

0.5(v^2) + 0.5r(v) - 6.4746 = 0 where gh = .66*9.81 = 6.4746

from here you can plug in your value for r and then treat v as the variable in the quadratic formula and find a value for it either using your calculator or the quadratic equation

*This answer is assuming you've figured out r somewhere else. I don't know how you would solve it without knowing r*