Ball height reached when rolling on inclines - rough or smooth

[Bonulo]

I've been presented with the following problems:

SITUATION:
A ball is rolling without slipping with velocity $v$ on a horizontal surface. It reaches an incline, which forms an angle $\theta$ with the horizontal. In which situation will the ball reach the highest point, when the incline has a rough surface, so the ball does natural roll or when the surface is completely smooth?
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Obviously, the difference is the lack of friction in the latter situation. The friction $f_r$ in the first situation makes the ball's rotational velocity drop, and the tangential component of the ball's weight $w$, $m g sin(\theta)$ in both situations slows down the ball's translational velocity.

I'm not sure whether the ball on the smooth surface will continue rolling (naturally), since there's no torque applied, since it was rolling naturally immediately before it met the incline.PROBLEM a)
Explain your answer with regards to conservation of energy.
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The equation is $U_1 + K_1 + W_[other] = U_2 + K_2$, and I can make it $K_1 + W_[other] = U_2$. But does the rolling friction do any work (when the ball rolls on the rough surface)? And how can the question of which ball reaches the biggest height be answered - by elmininating the rotational kinetic energy for the ball on the smooth surface, or by the work, if there is any?

PROBLEM b)
Explain your answer with regards to Newton's laws.
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I guess I have to use Newton's 2nd law here, and show that the existence of friction on the rough surface gives the ball linear acceleration up the incline, while slowing its rotation down.

Am I right? And if I'm not; am I far from the truth? :shy:

 

[Meir Achuz]

You are close to right.
No work is done since no force moves through a distance.
The static friction on the ball does no work.
(By the way, "rolling friction" means something else.)
By cons. of energy, the ball loses its rotational energy in the case of the surface with static friction, and will rise higher.
In terms of NII, draw a FBD of the ball and you will see that the plane with friction exerts an upward force (not acceleration) on the ball in stopping its rotation.

 

[Bonulo]

Yeah, I found out about that shortly after the posting.

So:

a) Rotational kinetic energy can only be utilized for forward linear motion if the surface is not smooth.

b) The friction exerts upward force, but it also exerts a torque. Does the torque affect the linear, tangential acceleration upwards?

 

[DaMastaofFisix]

As a mater of fact, yes it does. Using the conservation of energy method, the conclusion you came to was that as the ball rolls up the smooth incline, there was no torque provided by a friction force. therefore, none of the rotatioanl kinetic energy was converted into gravitational potential energy, menaning that only the translational kinetic enrgy was affected.

When considering the conservation of energy of a "rough" incline, the same setup applies. At first, all you have is rotational and translational kinetic energy. All that energy is converted into gravtitational potential energy atthetop of the hill.

So what dpes all of that mumbo jumbo mean? I means that the difference between the two scenarios is that the second equation has an extra term in it (the rotational kinetic energy), which means the that the final height has to be larger than in the first case.

As for Newtons laws, when you draw a free body diagram for your rolling ball, you will see that there are three forces: the component of the weight (mgsin(theita)), the Normal force provide by the plane and the frictional force (at least in the second case).

From this FBD, you should write your net force and net TORQUE equations for the x and y directions. Recognizing what forces produce a torque and so on, you can calculate the linear and angular accelerations. Kinda interseting and surprising to see what it all turns out.

 

[Bonulo]

Well, there is rotational kinetic energy in both cases - since the ball's rotating before hitting the incline. But it isn't converted into potential energy in the case of the smooth incline. I think.

Using Newton's laws I see that the rough surface has a positive linear acceleration that's $\mu m g cos(\theta)$, and an angular acceleration (from the friction) which, with the natural roll of the rough surface, gives a negative linear acceleration $a = -5/2 \mu g cos(\theta)$. The net linear acceleration of the rough surface ball is then $-g sin(\theta) + \mu g cos(\theta) - 5/2 \mu g cos(\theta) = -g sin(\theta) - 3/2 \mu g cos(\theta) = -g (sin(\theta) + 3/2 \mu cos(\theta))$. The net linear acceleration of the smooth surface ball is $- g sin(\theta)$. So the smooth surface ball has less acceleration downwards?

Or is something wrong?

 

[Doc Al]

One thing you are doing wrong is assuming that the friction force must equal $\mu mg \cos \theta$, but that's the maximum value of static friction, not the actual value (which will be less).

The way to analyze this is as follows. First identify the forces: (1) gravity ($mg$) acting down, (2) the normal force ($m g \cos\theta$), and (3) the unknown friction force (F) acting up the incline.

Then apply Newton's 2nd law for both translation and rotation. But don't forget the contraint that the ball rolls without slipping, which means that $a = \alpha R$. Now solve for the translational acceleration with and without friction. (You know the answer without friction: $a = g \sin \theta$.)

 

[Bonulo]

A possible solution (with N2)

Yeah, I forgot that. Then:

I calculate the two accelerations (linear and angular), put them in $a = \alpha R$, determine $f$ (the friction) thus solving for the linear acceleration:

SMOOTH SURFACE (no friction): $a = - g sin(\theta)$ .

ROUGH SURFACE (friction $f$): $a = - (g sin(\theta))/r$ .

Is that right?

 

[Doc Al]

SMOOTH SURFACE (no friction): $a = - g sin(\theta)$ .

Right.

ROUGH SURFACE (friction $f$): $a = - (g sin(\theta))/r$ .

I don't understand what this is. What's "r"? (If it's the radius of the ball, then the equation is dimensionally incorrect.)

If you assume a solid ball, you should be able to explicitly find the acceleration in terms of g.

 

[Bonulo]

I tried again, and got

$a = - 5/7 g sin(\theta)$

which is smaller than in the smooth situation.

 

[Doc Al]

That "natural roll" constraint allows you to relate the equations for linear and rotational dynamics, which enables you to solve for the two unknowns: friction and acceleration.

 

[Doc Al]

I tried again, and got

$a = - 5/7 g sin(\theta)$

which is smaller than in the smooth situation.

Now you've got it!

 

[Bonulo]

Yup ;) I'd merely choked on one of the equations.

And regarding the explanation using conservation of energy; I conclude that the largest U2 (for the greatest height) equals K1 - on the rough surface. Since there is a K2 (only the rotational kinetic energy) - on the smooth surface:

$U2 = K1 - K2$, $K2(rough) = 0 , K2(smooth) > 0$.

And - there is no work done by the friction? The textbook doesn't mention any in any equation or example.

 

[Doc Al]

And regarding the explanation using conservation of energy; I conclude that the largest U2 (for the greatest height) equals K1 - on the rough surface. Since there is a K2 (only the rotational kinetic energy) - on the smooth surface:

$U2 = K1 - K2$, $K2(rough) = 0 , K2(smooth) > 0$.

I think you understand it. The rolling ball has rotational and translational KE as it hits the incline. If the incline is rough, the rotational KE can be transformed into translational KE and thus into greater gravitational PE; if smooth, it cannot: the rotational KE remains rotational.

And - there is no work done by the friction? The textbook doesn't mention any in any equation or example.

Right. Static friction does no work. Mechanical energy is conserved in both cases. (If there was slipping between the surfaces, instead of rolling without slipping, then dissipative work would be done by the kinetic friction and mechanical energy would not be conserved.)

 

[Bonulo]

Good, thanks Doc Al, Meir Achuz and DaMastaofFisix!