Asolid uniform cylinder of mass m and radius r is projected up an incline of angle q. it rolls without slipping from an initial speed of v0 of the center of mass. what distance s does the center of the cylinder travel before it starts to fall back?

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)

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answer: (431,2)

step-by-step explanation: