Answer:
s = 0.75*v₀² / (g*Sin q)
Explanation:
Loss in rotational plus the translational kinetic energy is equal to the gain in the gravitational potential energy of the cylinder.
As it does not slip, total mechanical energy will be conserved.
Energy at position 1 is:
E₀ = Kt + Kr
As
Kt = 0.5*M*v₀²
and
Kr = 0.5*Icm*ω²
where
Icm = 0.5*M*R²
and
ω = V / R
we have
E₀ = (0.5*M*v₀²) + (0.5*(0.5*M*R²)*(v₀² / R²)) = 0.75*M*v₀²
Energy at position 2 is
E₁ = M*g*h
where
h = s*Sin q ⇒ E₁ = M*g*s*Sin q
Finally we apply:
E₀ = E₁
⇒ 0.75*M*v₀² = M*g*s*Sin q
⇒ s = 0.75*v₀² / (g*Sin q)
