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In which case will the kinetic energy be more after reaching the bottom? The shaded surface have friction and other are smooth.

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I thought about these questions:

  • Does the answer depend on whether the wedge is fixed or not?
  • Will there be any slipping?
  • What is the work done by friction?
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closed as off-topic by Kyle Kanos, John Rennie, Jon Custer, sammy gerbil, ZeroTheHero Apr 2 '18 at 14:21

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  • $\begingroup$ As it's in rotational motion , the friction would not be dissipative I think .... Moreover , if both the frictioned regions are equal , then kinetic energy would be same ... $\endgroup$ – user182687 Mar 24 '18 at 11:50
  • $\begingroup$ If there is slip, then friction would do work. I just can't figure out where and how. Because, the answer given is not equal. $\endgroup$ – Ice Inkberry Mar 24 '18 at 11:57
  • $\begingroup$ In the frictioned part , there is rotational kinetic energy and in frictionless part there is translational kinetic energy ... The first ball loses some translational energy as rotational , but eventually obtains it later ... But the second ball experiences rotational motion at the end , which is not recovered ... $\endgroup$ – user182687 Mar 24 '18 at 12:07
  • $\begingroup$ So , I think first ball has more kinetic energy ... $\endgroup$ – user182687 Mar 24 '18 at 12:07
  • $\begingroup$ Didn't get 'The second ball experiences motion at end, which is not recovered' $\endgroup$ – Ice Inkberry Mar 24 '18 at 12:10
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Assume the wedge does not move ie is anchored to the ground.

For case (i) make the assumption that there is no slipping so no work is done by the frictional force and the decrease in gravitational potential energy is equal to the increase in translational and rotational kinetic energy.

For case (ii) one cannot make the no slipping assumption for all time and so some of the mechanical energy will become heat. This means that the decrease in gravitational potential energy will be greater than the gain in translational and rotational energy.
The reason that there must be slipping at some stage in case (ii) is that when the object reaches $B$ it is not rotating and only has translational motion.
The object cannot instantaneously rotate fast enough for the no slipping condition to be satisfied.
So there must be a period of time during which the object is increasing its rotational motion whilst at the same time slipping.

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for finding speed of the object we use

v= {2gh/ (1+I/mr²)}½

where I is the rotational inertia of the rolling object. h is the height of the wedge.

as wedge is like right angled triangle using pythagorus theorem (AB+BC)² = h² + L² ; L: length of the wedge. use h from this equation to the above and then the k.E for translation motion as mv²/2 and rotational k.E as I r² v²/2 then sum them for total.

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