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 it 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? How high would the marble go if both sides were as rough as the left side?please explain why and show calculations.

A. Less than h; h

B. Less than h; less than h

C. h; h

D. More than h; h

E. more than h; less than h

Answer 1

use law of conservation of energy,

initial P.E=total K.E at bottom

m*g*h=1/2*m*v^2+1/2*I*w^2

m*g*h=1/2*m*v^2+1/2*(2/5*m*R^2)*(v/R)^2

g*h=1/2*v^2+1/5*m*v^2

g*h=7/10*v^2

===> v^2=10/7*g*h

speed of the marble on reaching the ground is,

v=sqrt(10/7*g*h)

now,

total energy of the marble at the upward sliding is,

E=K.E+P.E

E=1/2*m*v^2+0

E=1/2*m*(10*g*h/7)

E=5/7*m*g*h

if the marble reaches the height h'

P.E=total energy

m*g*h'=5/7*m*g*h

h'=5/7*h

===> h'< h option B is correct