Question

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?

Answer 1

Step-by-step explanation:

a ) When the marble goes down , its potential energy is converted into both linear as well as rotational kinetic energy . So its total potential energy gets distributed between two components . Hence, the component of linear kinetic energy which helps it to go up the right side wall gets reduced . As a result of it the height up to which it ascends on the right side smooth wall is comparatively less than h . At the top position on the right side smooth wall , it keeps on rotating only having no linear upward motion there.

b ) When both sides are rough , marble will not keep rotating on the right side at the top , because friction on the right wall will stop it rotating . Hence it will be able to go upto the same height h as the left wall in this case.

c ) In case of absence of friction , it bears some of its energy in the form of rotational energy so height achieved by it becomes less. In case of friction , its rotational energy gets converted into linear kinetic energy , thus increasing the height achieved by it . In the whole process , it must be borne in mind that friction i not playing the role of dissipative force here because it is a case of perfect rolling and not slipping which causes dissipation.