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There are two fixed inclined plane as shown above. In situation (A), there are frictions on the left side. The right side of the inclined plane is frictionless. In situation (B), the right side of the inclied plane are frictionless. There is friction on the left side. (Like the picture below)

Both the situation A and B, A rigid ball starts from the left side of the inclined plane. The initial velocity is 0.

If you calculate, the maximum height that the ball goes in situation A and B is both $ \frac{h}{1+\beta}$ where the rotational inertia of the ball is $\beta mr^2$. (It does not depends on the friction constant $\mu$) The fact that the height is the same at A and B really interests me.

With a little calculation, you can figure it out but I want qualitative reason why the maximum heights are the same at A and B.


Add) Here are the derivation I did.

(A): The energy of the ball conserves. Let $v$ be the velocity of the ball at the lowest point (with friction). $ mgh=\frac{1}{2}\beta m r^2\omega^2+\frac{1}{2}mv^2=mgh_f+\frac{1}{2}\beta m r^2\omega^2 $ with the fact that $v=rw$, $h_f=\frac{h}{1+\beta}$. For those of you who are confused about the situation, Starting from the frictionless area, no torque applies to the ball so the ball spins with a certain angular velocity. That's why the $h_f$ is smaller than $h$. It's because part of the initial energy turns into rotational energy.

(B): The ball looses energy while entering the friction area because the ball wasn't rolling. the velocity of the ball at the frictionless area is $v=\sqrt{2gh}$. Let the point of the torque lies anywhere on the lowest point. And since friction doesn't apply and torque (friction vector passes the torque messing point), angular momentum of the ball conserves between frictionless point and the friction point(where the ball rolls without slipping). $mr\sqrt{2gh}=\beta mr^2 \omega_{friction}+rmv_{friction}$. so, $v_{friction}=\frac{\sqrt{2gh}}{1+\beta}$. If you use energy conservation between the friction point and the final ball's maximum height $h_f=\frac{h}{1+\beta}$

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I have added to my answer to try and give a better explanation.

If the ball is rolling without slipping the frictional force does no work as there is no relative movement at the point of contact between the ball and the surface.

enter image description here

For the left-hand situation mechanical energy is conserved.
The ball starts from rest at $A$ and gains both rotational and translational kinetic energy as it loses gravitational potential energy with no work done by friction as there is no slipping.
At $B$, $mgh = \frac 12 mv^2 + \frac 12 \beta mr^2 \omega^2$ and $v=r\omega$.
From $B$ to $C$ the frictional force does no work and so the ball continues to $D$ at constant total (translational and rotational) kinetic energy, indeed the frictional force is zero in this phase of the balls motion.
From $D$ to $E$ the rotational speed of the ball cannot change as there is no frictional or any any other force about its centre of mass to apply a torque on it. Thus the only change in mechanical energy is that of the translational kinetic energy being converted into gravitational potential energy.
At $E$ the ball ceases any translational motion but is still rotating about its centre of mass and so has rotational kinetic energy. Thus $mgh' = \frac 12 mv^2 \Rightarrow h'=\frac {h}{1+ \beta}$

The situation is different for the right-hand situation where the is a region where mechanical energy is not conserved.
From $A$ to $F$ the ball falls gaining only translational kinetic energy, $mgh = \frac 12 m u^2$ and note that $u>v$.
From $F$ to $G$ the ball is not rotating and has constant speed, $u$, and constant translational kinetic energy, $\frac 12 mu^2$.
When the ball arrives at $G$ the ball is not rotating and so there must be relative movement between the ball and the surface so kinetic friction, $f_{\rm k}= \mu_{\rm k}mg$, must be acting.
The kinetic friction will be in the opposite direction of the motion of the ball to reduce its translational speed whilst at the same time apply a torque about the centre of mass of the ball to increase the rotational speed of the ball.
This will continue until the no slip condition is satisfied with $u' = r\omega'$.
So after $G$ the kinetic frictional force is doing reducing the mechanical energy of the ball.
If the time to reach the no slipping condition is $t$ then using the ideas of impulsive force and impulsive torque $f_{\rm k}\,t = mu-mu'$ and $f_{\rm k}\,r\,t = \beta mr^2\omega -0'\Rightarrow u' = \frac {u}{1+\beta}$.
After reaching $H$ the ball gains gravitational potential energy and loses both rotational and translational kinetic energy until it reaches a height $h''$ where is stops both translational and rotational motion.
$mgh'' = \frac 12 \beta mr^2 \omega'^2 + \frac 12 m u'^2$ with $u' = r \omega'$ which then produces the result $h'' = \frac{h}{1+\beta}$, the same height as in the left-hand situation.

So the mechanical energy lost in the right-hand situation is equal to the rotational kinetic energy the ball had when it stopped translational motion in the left-hand situation.

Static friction does play a part in both situations but in the right-hand situation there is also kinetic friction acting for a part of the motion.
It is interesting to note that the magnitude of the kinetic frictional force does not affect the final result; if it is large the time taken from pure translation to translation and rotation is small and that occurs over a shorter distance as compared with the kinetic frictional force being small so that the transition time is now larger and the distance over which the transition occurs is also larger.
However the work done by the kinetic frictional force, frictional force $\times$ distance during which the transition occurs, stay the same!

In some ways I would compare this with the charging of a capacitor in a battery, resistor and capacitor series circuit, where the final energy stored in the capacitor does not depend on the resistance of the resistor.

How can friction do no work in case of pure rolling?

Work done by a friction in rolling

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  • $\begingroup$ I agree with what you say, but if the ball is rolling without slipping for both A and B then isn't mechanical energy conserved, which would mean $h_{final}= h_{initial}$ not $h_{final}\lt h_{initial}$ as the OP states? $\endgroup$
    – Bob D
    Nov 4, 2022 at 19:50
  • $\begingroup$ I haven't gone through your answer in detail, but I'm trying to resolve two statements. First you make the statement "For the left-hand situation mechanical energy is conserved". To me that means the ball rolls down without slipping and the final height of the ball would equal the original height $h$. For the right side situation you say slipping occurs while rolling resulting in kinetic friction losses. To me that means mechanical energy is not conserved. Yet you state " which then produces the result $h'' = \frac{h}{1+\beta}$, the same height as in the left-hand situation." $\endgroup$
    – Bob D
    Nov 5, 2022 at 14:24
  • $\begingroup$ How can the ball reach the same height on the right side as the left side if mechanical energy is not conserved on the right side but is on the left side? What am I missing? $\endgroup$
    – Bob D
    Nov 5, 2022 at 14:24
  • $\begingroup$ @BobD - to quote from my answer you are missing the point that the mechanical energy lost in the right-hand situation is equal to the rotational kinetic energy the ball had when it stopped translational motion in the left-hand situation and still had rotation kinetic energy. $\endgroup$
    – Farcher
    Nov 5, 2022 at 14:26
  • $\begingroup$ Let's take this one step at a time. You are saying that mechanical energy is conserved on the left side, correct? $\endgroup$
    – Bob D
    Nov 5, 2022 at 14:29
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It does not depends on the friction constant $\mu$

It most certainly does, particularly in the second scenario. What is being assumed in that scenario is that friction suddenly becomes so high that the ball instantaneously transitions from sliding with zero angular velocity to rolling without slipping while conserving energy.

The only way to obtain $h_\text{final} = \frac{h_\text{initial}}{1+\beta}$ is to assume that the ball does not transfer energy to the surface. This means the total energy (translational + rotational + gravitational) of the ball is conserved. With the assumption that the ball's total energy is conserved, the total energy (translational plus rotational) when the ball reaches the bottom of the ramp by rolling without slipping in first scenario is the same as the total energy of the ball at the bottom ramp in the second scenario when it starts to roll up the ramp without slipping. It shouldn't be surprising that both scenarios yield the same final height.

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  • $\begingroup$ "The only way to obtain $h_\text{final} = \frac{h_\text{initial}}{1+\beta}$ is to assume that the ball does not transfer energy to the surface." It seems to me the only way the final height is less than the initial height is to assume the ball DOES transfer energy to the surface in the form of friction heating. Otherwise total mechanical energy would be conserved and $h_{final}=h_{initial}$ since $\Delta KE$ is zero (ball begins and ends at rest). $\endgroup$
    – Bob D
    Nov 4, 2022 at 19:16
  • $\begingroup$ @BobD but the ball's speed on the surface is determined by the translation. And the ball is not doing pure translation in some cases. $\endgroup$ Nov 4, 2022 at 21:15
  • $\begingroup$ @BobD in the second situation the ball will not go same as initial height because the translation speed will be reduced on friction part. $\endgroup$ Nov 4, 2022 at 21:30
  • $\begingroup$ @Teaislife When you refer to the "second part" do you mean diagram B? $\endgroup$
    – Bob D
    Nov 4, 2022 at 21:34
  • $\begingroup$ @BobD yes...... And in diagram A the ball already has less translation speed and continues with that speed with rotation. If everything was frictionless the ball will go to same initial height. $\endgroup$ Nov 4, 2022 at 21:35
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Here is what happens

In the first diagram, the potential energy is converted to rotational plus translational.

U=R+T

This translational velocity is lesser than a pure translational velocity. So the ball will have lesser potential energy in the other end. But the ball will keep rotating.

U=R+T=R+U'

In the second diagram, the ball does a pure translation and then suddenly rotates with a reduced translation speed on touching the friction part. The friction produces an angular momentum which was not there before. So some energy must be lost to make the ball rotate and to get equal height to first diagram, the rotational energy in the second diagram must be dissipated and all the lost energy must be equal to the rotational energy at the end of the first diagram.

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