I tried using this approach to solve another problem, however I couldn't find anything no matter how hard I tried. I am not really familiar with problems of this sort and mechanics in general. Because I couldn't find a solution for so long, I got desperate and asked chatgpt but all it suggested was a constant function. I hope I have overlooked something, please tell me if you have ever heard of something similar.
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$\begingroup$ Is it allowed for a gravitational potential energy of ball be dissipated by inelastic collisions,- i.e. by heat, deformation and so on ? Or in this problem only pure rolling (without loosing contact with ground) is allowed ? $\endgroup$– Agnius VasiliauskasCommented Nov 27 at 21:57
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$\begingroup$ How could that work? When the ball starts rolling it would have to accelerate instantly from 0 to v. $\endgroup$– PM 2RingCommented Nov 28 at 11:14
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6 Answers
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There is only the constant curve. This is because energy is conserved
$$E = E_{\text{kinetic}} + E_{\text{potential}} = \text{const}.$$
so any curve that a ball rolls down (i.e. not a constant curve) will result in less potential energy and thus more kinetic energy for the ball. this means the velocity of the ball would increase
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Lorenz'z answer is the correct one if the only force operating on the ball is gravity. If you add friction then there will instead be a 'curve' consisting a path of constant slope.
The only way to get a curved path would be if the gravitational force and/or the coefficient of friction varied with position.
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$\begingroup$ I take "rolling" in the question to mean "rolling without slipping," in which case friction won't be dissipating any mechanical energy. $\endgroup$– d_bCommented Nov 27 at 20:38
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$\begingroup$ My answer would be valid for any type of friction including air resistance, slipping, 'stiction' etc. But I agree, no friction = no slope. $\endgroup$– PenguinoCommented Nov 27 at 20:44
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This depends on the amount of friction you have. you need to adjust the slope such that the friction forces $F_\text{fric}$ force cancels the slope downforce $F_\text{slope}$ at the desired velocity.
If the friction coefficient is constant, your curve will be a straight line.
In the idealized case of no friction, your curve would just be a horizontal line. This is because constant velocity needs no acceleration.
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Without friction, the curve is a horizontal line.
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If gravitational potential energy is not allowed to be dissipated and energy conservation must be obeyed,- then answer is NO, such constant speed curve does not exist. But, if potential energy dissipation by any means IS allowed, then couple of schemes exist how constant speed curve can be accomplished.
One of them,- dissipation of gravitational potential energy with inelastic ball collisions with ground. In this case this potential energy will be converted into heat/ground or ball deformation and then ball can continue on the same or approximate speed. Typical staircase curve will do :
Blue line is possible trajectory of a ball going downstairs.
P.S. Staircase material or ball material must be chosen so that correct amount of potential gravitational energy would be dissipated per staircase step, i.e. $\delta E_{step}=mgh_{step}$
answered Nov 28 at 8:16
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A freebody diagram of the ball will have three forces, ideally in a vacuum. Weight, normal and friction forces. The difference in direction of the normal force and weight add to a resultant force. The only way the two will add to zero is if the slope is zero. Rolling down means the slope is not zero. Any slope at all will have a force and an associated acceleration.
The rolling friction and mass moment of inertia will retard that acceleration, but not limit maximum speed. Potential energy converts to rotational energy.
If in a atmosphere, fluid, gas, or liquid, there will be a terminal velocity associated with the slope, even if strait down. A curve will modified the slope so acceleration forces won't be constant, but always present.
An alternate thought will be that choice of slope and viscosity could potentially lead to constant acceleration. That is constrained by terminal velocity too.
