Hi there!
We know that:
Ei = Ef
At the bottom of the incline, we have both rotational and translational kinetic energy. Recall:
Translational KE = 1/2mv²
Rotational KE = 1/2Iw²
We can begin by deriving an equation for the final energy of the sphere.
Moment of Inertia of a hollow sphere= 2/3mR²
[tex]KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2(\frac{v^2}{R^2}))[/tex]
Simplify:
[tex]KE_{total} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2[/tex]
Since Ei = Ef, and the initial energy is simply PE = mgh, thus:
mgh = 5/6mv²
Cancel out 'm':
gh = 5/6v²
We are given the angular velocity, so we can convert to velocity using:
v = ωr
v = 5.88(.68) ≈ 4.00 m/s
h = (5/6v²)/g
h = (5/6(4²))/g = 1.359 m
