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.
My doubt is, once the ball enters to the smooth surface after rolling thro' the rough surface, what happens to the Rotational Kinetic energy?? I mean will it remain the same or it will be zero since there is no torque on the marble....
I think rotational K.E will convert in translational K.E and thus it will start slide over the surface and will move same distance
the rotational kinetic energy shall remain the same....the ball would be spinning with the same angular velocity when it reaches the top of the other end of the bowl....the only difference would be that it would now rise to a lower height than it was released from.
So it means "First the Potential energy is converted into KE and reaches the bottom of the bowl then since there is no torque on the marble it uses the KE to rotate the marble in the smooth surface and hence reaches lower height"...... Is my understanding correct??
no...the potential energy is first converted into translational and rotational kinetic energy as it reaches the bottom...then the translational kinetic energy is converted to potential energy as it goes to the other side...it reaches a lower height as the ball has not given up its rotational kinetic energy..and resumes the spinning