Question

em 8 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 botom of the bowl. The left half of the bow Part A is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. How far up the smooth side will the marble go, measured vertically from the bottom? Express your answer in terms of h Submit Request Answer Part B How high would the marble go if both sides were as rough as the left side? Express your answer in terms of h Submit Provide Feedback

Answer 1

Using conservation of energy for the system. The speed of the ball at the bottom Rightwardsdoublearrow choosing bottom as reference. 1/2 mv^2 + 1/2 Iw_A^2 + mgh = 1/2 mv^2_B + 1/2 w^2_B + 0 v_A = w_A = 0 initially mgh = 1/2 mv^2_B + 1/2 Iw62_B V_B = rw_B where r = radius of the marble. Rolls without slipping. I_solid sphere = I_marble = 2/5 mr^2. Mgh = 1/2 mv^2_B + 1/2 times 2/5 middot mr^2 (V_B/r) ^2 gh = 1/2 V^2_B or V^2_B = 10/7 gh (speed) ^2 at the bottom of the bowl. Since no friction acts on the right nalf of the bowl. Hence torque doesn't act on the marble, so its rotational kinetic energy doesn't using energy conservation. 1/2 mv^2_B + 1/2w^2_B rotational kinetic energy doesn't change = 1/2 Iw^2_B + mgh^1 V^2_B = 70/7 gh Therefore 1/2 m(10/7 gh) = mgh^1 h^1 = s/7 n height to which marble rises from bottom of the bowl. Part B