Help!! Rolling Sphere up a ramp problem...?

The ball's total kinetic energy is the sum of its translational and rotational kinetic energies:

KE = ½mv² + ½Iω²

I = 2/5mr² and ω = v/r

KE = ½mv² + ½(2/5mr²)(v/r)²

Simplify (radius cancels out)

KE = ½mv² + 1/5mv²

KE = (7/10)mv²

Now set up an energy initial = energy final equation.

KEi + GPEi = KEf + GPEf

Input the values for each of these. We'll call GPEi 0.

(7/10)m(V_B)² = (7/10)m(V_T)² + mgh

From here you just solve for V_B

Yes, you should use conservation of energy.

Moving up the ramp, the sphere gains potential energy. This gain has to equal its loss of kinetic energy.

A rolling sphere has two types of kinetic energy: part is based on its linear speed, and part is based on its rotation around its center of mass.

KE = 1/2 m v^2 + 1/2 I ω^2

Remember that ω = v/r. So this becomes

KE = 1/2 m v^2 + 1/2 (2/5 mr^2) (v/r)^2 = 7/10 m v^2

This kinetic energy uses v = V_T at the top of the ramp and v = V_B at the bottom.

The change in potential energy is mgh.

Conservation of momentum thus gives the following equation:

7/10 m V_B^2 = mgh + 7/10 m V_T^2

Solve for V_B

V_B = √(10/7 gh + V_T^2)