A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil.(a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

a)  h'= 5/7 h , b)   h ’= h

Explanation:

Let's use energy conservation for this exercise

Starting point. Upper left side

         Em₀ = mg h

Final point. Lower left side

        Emf = K = ½ m v² + ½ I w²

        Em₀ = emf

        mgh = ½ m v² + ½ I w²

Angular and linear velocity are related

           v = r w

           w = v / r

The moment of inertia of the marble that we take as a solid sphere is

           I = 2/5 m r²

We substitute

          m g h = ½ m v² + ½ 2/5 m r² (v / r)²

          g h = ½ v2 (1 + 2/5)

         v = √(g h 10/7)

This is the speed at the bottom of the bowl

Now let's apply energy conservation to the right side

a) right side if rubbing

             Em₀ = K

              Emf = U = mg h’

             ½ m v² = mg h’

              h’= ½ (g h 10/7) / g

              h'= 5/7 h

b) right side with rubbing

             Em₀ = K

             Emf = K + U = -½ I w² + m g h

             Emf = -½ 2/5 m r² v² / r² + m gh

            Em₀ = emf

            ½ v² = -1/5 v² + g h’

            h’= (1/2 +1/5) (gh 10/7) / g

            h ’= h

c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher

did yall get answers for this yet