Question

15) A uniform marble rolls down a symmetrical 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

a)applying energy conservation at the points top left and the bottom of the bowl

(work done due to friction will be 0 since there is no relative slipping between the bowl and marble)

$mgh = mv^{2}/2 + I\omega ^{2}/2 = mv^{2}/2 + (2/5*mr^{2}*v^{2}/r^{2}*1/2) =7mv^{2}/10$..............eq1

again apply energy conservation at the points top right and the bottom

$mv^{2}/2 + I\omega ^{2}/2 =mgh_{topright} + I\omega ^{2}/2$............eq2 (since there is no friction so there is no torque about center of mass so angular velocity will remain same)

htopright 5/1/7

b) applying enery conservation at top left point and the bottom left point

(work done due to friction will be 0 since there is no relative slipping between the bowl and marble and kinetic energy is also 0 at the initial and final points)

mghtopleft = mghtopright

htopleft - htonriaht

c) when the only side was rough only translational energy at the bottom was contributing to increase the potential energy while when both sides were rough both translational and rotational energy were contributing to the cause.