Sciencemadness Discussion Board - Brain teaser: rolling ball with friction

Brain teaser: rolling ball with friction

blogfast25 - 8-3-2016 at 11:57

A solid sphere of R=0.1 m is made to rotate counter-clockwise at 150 rpm about an axis parallel to the y-axis and through its centre of gravity. It’s then launched at 1 m/s on a smooth horizontal xy-plane in the x-direction. The kinetic coefficient of friction between the sphere and ball surfaces is 0.01. Ignore air-drag.

What’s the final horizontal velocity of the sphere?

aga - 8-3-2016 at 12:37

0

woelen - 8-3-2016 at 12:47

Define dynamic variables:
vc : Velocity of center point, initially this equals 1 m/s
w : angular velocity in counter clock direction, initially this is 2*pi*150/60 radians per second
vf : Velocity of contact point on sphere relative to the floor
Ff : Force due to friction with floor
T: Torque due to friction with floor

Parameters:
r : radius of sphere: 0.1 m
mu : 0.01
M : not given, not needed for final answer
J : moment of inertia, for sphere with uniform density this is 2*M*r*r/5. This is an omission in the brain teaser. You did not state that density is uniform, I assume that this is the case.

Velocity of contact point: vf = vc + w*r
Force on contact point: Ff = -mu * vf = -mu * (vc + w*r)

dvc/dt = Ff/M
dw/dt = T / J = r*Ff/J

Using equation for Ff:

dvc/dt = -mu/M * (vc + w*r)
dw/dt = -mu*r/J * (vc + w*r)

Now we have two coupled differential equations with initial conditions vc = 1 m/s and w = 2*pi*150/60 rad/s. This system can be solved easily.

Solving this system is trivial from a conceptual point of view, but involves a lot of tedious algebra. I do not give the derivation of the solution, I simply present the numerical results for the given parameters and initial values. I substituted some M (M = 1 is OK) and found the following:

final velocity of center of mass at 6 digit acuracy is 0.265487 m/s
final angular velocity at 6 digit acuracy is -2.65487 rad/s

If you change M, then the solution scales in time. E.g. when M is made equal to 10, then the same solution is obtained, only 10 times slower. For the final value, however, this does not matter. The final value is reached asymptotically.

[Edited on 8-3-16 by woelen]

aga - 8-3-2016 at 13:08

Awe. Just overwhelmed with awe.

blogfast25 - 8-3-2016 at 13:16

Quote: Originally posted by woelen

I. You did not state that density is uniform, I assume that this is the case.

II. final velocity of center of mass at 6 digit acuracy is 0.265487 m/s
final angular velocity at 6 digit acuracy is -2.65487 rad/s

III. If you change M, then the solution scales in time. E.g. when M is made equal to 10, then the same solution is obtained, only 10 times slower. For the final value, however, this does not matter. The final value is reached asymptotically.

I. A bit harsh, I feel. What else is 'solid sphere' really supposed to mean here?

II. Correct.

III. Incorrect. Full derivation shows it to be mass invariant. That's where that "a lot of tedious algebra" comes into it!

I'll wait with my derivation, in case others want to try.

woelen - 8-3-2016 at 13:58

Quote:

A bit harsh, I feel. What else is 'solid sphere' really supposed to mean here?

This is not meant to be harsh, if it is perceived as harsh by you, I want to apologize. I only wanted the problem to be stated very precisely. A solid sphere does not need to be uniform.

Quote:

III. Incorrect. Full derivation shows it to be mass invariant. That's where that "a lot of tedious algebra" comes into it!

No it is not mass invariant, I am correct. Only the final value is mass invariant, the road towards the final value is not, its "time constant" is proportional to the mass. E.g. if the mass is 10 times as large, then it takes 10 times as much time to reach the asymptotic value within e.g. 1%.
You can also imagine this without any computation. Suppose an experiment with a sphere of 1 kg. As soon as it touches the floor, there is a certain force -mu*vf. This force does not depend on the mass. Now imagine a sphere of one tonne (made of compressed star matter

with the same size). The force still is the same -mu*vf, but the mass is 1000 times as large, and its moment of inertia also is 1000 times as large. So, the same force only has 1/1000 of the effect on a sphere of 1 kg in terms of deceleration and angular deceleration. With this little thought experiment you can see that the system is not mass invariant, its response is slower when the mass is larger. Derivation of the solution of the differential equation shows that there is a time constant (in the exponent of the solution), which is proportional to M.

[Edited on 8-3-16 by woelen]

blogfast25 - 8-3-2016 at 14:36

Quote: Originally posted by woelen

Only the final value is mass invariant,

Actually, that's not what I'm finding:

$$F_F=mg\mu_k$$
$$ma=-mg\mu_k$$
$$\frac{dv}{dt}=-g\mu_k$$
$$\int_{v_0}^v dv=\int_{0}^t(-g\mu_k)dt$$
$$v=v_0-g\mu_kt$$
$$\tau=mg\mu_kR$$

$$I\alpha=-\tau$$
$$I\alpha=-mg\mu_kR$$

$$\frac{d\omega}{dt}=-\frac{mg\mu_kR}{I}$$
$$\int_{\omega_0}^{\omega}d\omega=\int_0^t (-mg\mu_kR)dt$$

$$\omega=\omega_0-\frac{mg\mu_kR}{I}t$$

$$v=-\omega R$$

$$v_0-g\mu_kt=-(\omega_0-\frac{mg\mu_kR}{I}t)R$$

$$I=\frac{2mR^2}{5}$$

$$t=\frac{\omega_0R+v_0}{3.5g\mu_k}$$

$$\omega=2\pi\frac{150}{60}=15.7\:\mathrm{s^{-1}}, v_0=1\:\mathrm{m/s}, \mu_k=0.01$$

$$t=7.49\:\mathrm{s}$$

$$v=0.266\:\mathrm{m/s}$$

[Edited on 8-3-2016 by blogfast25]

wg48 - 8-3-2016 at 22:21

Here is my attempt.

The mass and coefficient of friction are not required as their ratio only effects the time to rolling. (assume the mass is unity)

Calculate the initial forward and rotational momentums.
The friction changes them in the ratio of 1 to pi/5 untill they are in the ratio of 1 to pi/5 (rolling). At that ratio the forward momentum gives the final velocity.

If initially the forward momentum is defined as positive and the rotational momentum is considered negative the friction will decrease the forward momentum and increase the rotational momentum.

[Edited on 9-3-2016 by wg48]

woelen - 8-3-2016 at 23:39

Ah, I see the difference between my outcome and your outcome. The final velocity is the same, but yours indeed is mass-invariant, also the path towards the final velocity. The solution to your problem is simpler, because the force is constant, as long as the sphere is slipping.

I had a problem in robotics in mind, with a roller, touching a surface. I interpreted the coefficient mu as a viscous friction: Ff = -mu * v, where v is the difference in velocity between the contact point on the roller and the velocity of the contact point on the surface, which is touched. You use mu to model the pushing of the sphere on the floor. As long as v is not equal to 0, there is slipping and then there is friction.

In your solution, the problem indeed is mass-invariant, again easily verifiable with a thought experiment. If the mass becomes tenfold, then the weight presses 10 times as hard on the surface and the force (and accordingly torque) also becomes tenfold and hence the net-effect in terms of acceleration and angular acceleration remains the same.
In my solution, the problem is not mass-invariant. Viscous friction does not depend on mass.

Indeed, I think that the term "kinetic friction coefficient" stands for your interpretation, although I also have seen my interpretation. This is a problem of wording and sometimes of people's background. But formally, your interpretation is the correct one.

A nice exercise would be to make the model even more realistic and take both forms of friction into account. In that case, Ff = -mu*m*g - nu*vc, where nu is the viscous friction coefficient and vc is the velocity of the contact point on the sphere, relative to the floor. In that case the solution is not mass invariant anymore, but also not simply time scalable as in my solution.
In robotics, the term -mu*m*g normally need not be taken into account. The weight of the roller is carried by the construction and touching other objects is done with minimal normal force. In that scenario, viscous friction is the dominant force, especially if the material is lubricated.

wg48 - 9-3-2016 at 02:29

Yes woelen I thought dynamic friction was an unusual term in that type of problem so I looked it up. I think it must be new school.

[Edited on 9-3-2016 by wg48]

woelen - 9-3-2016 at 03:00

I looked up the different words, used for different types of friction. In the english-speaking world the following terms seem to be used nowadays:

- kinetic friction: This depends on the normal force on the surface on which an object slides (not rolling). This is blogfast25's mu. An example of a website explaining this: http://www.softschools.com/formulas/physics/kinetic_friction...

- viscous friction:This is a form of friction where the friction force can be written as Ff = f(v), where f(v) is a continuous function with the property f(v) = -f(-v). In many cases, a simple model is used of the form F = -mu * v, where mu is the viscous friction coefficient and v is the velocity difference between the sliding surfaces. Sometimes, unfortunately, this type of friction also is called kinetic friction, but formally this is wrong. Here was my difference in the solution of blogfast25's problem.

- stiction, a.k.a. static friction: This type of friction usually is equal to the applied force on an object, until some breakdown value is reached. When the object moves, the stiction suddenly drops to zero and then viscous friction or kinetic friction (or a combination of these) sets in. Stiction is a nasty type of friction, both from a modelling point of view, and also in real constructions. The breakdown force in many situations is proportional to the normal force. Stiction also sometimes is confused with kinetic friction, but again, formally this is wrong.

blogfast25 - 9-3-2016 at 06:45

$$\mu_k$$

... is of course an idealisation: actual friction force then depends only on Normal force and kinetic friction coefficient, not on relative speed between the sliding surfaces. It's very commonly used in simple, dynamic friction problems but maybe I hang out at physics forums too much...

wg48 - 10-3-2016 at 04:19

I had difficulty determining how the friction force would accelerate the sphere ie how it would be shared between the linear inertia and rotational inertia. I ultimately used how rolling inertia is derived on the assumption that was correct.

I checked the other two solutions given but can not see how it was done in those cases. Can you explain please guys.

woelen - 10-3-2016 at 04:52

There is a force Ff at the contact point of the sphere/floor. It acts on the sphere in a direction, parallel to the floor, along the x-direction. This force Ff is equal to -mu*m*g for the kinetic friction model, used by blogfast25 or equal to -nu*v for the viscous friction model I used. In practice a combination of both effect is even more realistic.

The force Ff may be considered as if acting on the center of gravity of the sphere, although it acts on the outside of the sphere. This leads to a linear acceleration a = Ff/m along the x-direction.

The fact that the force Ff does not act on the center of rotation is taken into account by modeling a torque as well. The distance between the center of rotation and the point, where the force acts on the sphere equals R (in this problem the center of rotation is the same as the center of the sphere and the same as the center of gravity, but in general this need not be the case). The torque equals R*Ff and this leads to a rate of change of angular velocity, a.k.a. angular acceleration, which equals Ff/J, where J is the so-called moment of inertia. For a solid uniform density sphere, rotating around its center of gravity, this J equals 2*m*R*R/5.

Now we have acceleration and we have angular acceleration in terms of Ff.
Ff itself is a constant in blogfast25's case of kinetic friction, it depends on the velocity of the contact point relative to the floor in case of viscous friction. From here, it is a matter of solving a differential equation. In blogfast25's case, the solution has a non-homogeneous linear form y = a+bx, up until the time where the relative velocity of the contact point to the floor becomes 0 (rolling of sphere, no slipping/sliding). In my case of viscous friction, the solution has an asymptotic form a + b*(1-exp(-t/tau)). For time going to infinity, the solution moves towards a constant (rolling of sphere, no slipping/sliding).

I hope that this explanation helps. In fact, the problem is decomposed in two sub-problems, one solving the problem of deriving the equation for velocity and the other solving the problem of deriving the equation for angular velocity. The crux is in the determination of net force, acting on the body, and determination of net torque acting on the body.

wg48 - 10-3-2016 at 11:20

Quote: Originally posted by woelen

snip

The force Ff may be considered as if acting on the center of gravity of the sphere, although it acts on the outside of the sphere. This leads to a linear acceleration a = Ff/m along the x-direction.

The fact that the force Ff does not act on the center of rotation is taken into account by modeling a torque as well. The distance between the center of rotation and the point, where the force acts on the sphere equals R (in this problem the center of rotation is the same as the center of the sphere and the same as the center of gravity, but in general this need not be the case). The torque equals R*Ff and this leads to a rate of change of angular velocity, a.k.a. angular acceleration, which equals Ff/J, where J is the so-called moment of inertia. For a solid uniform density sphere, rotating around its center of gravity, this J equals 2*m*R*R/5.

Snip
.

I don't think you can consider the force to act at the center.
But it does depend on what's you mean by that.

The best I could come up with is the sphere is equivalent to a dumb bell of length 2R with each mass (point) half the rotation inertia. The axis of the dumb bell is normal to the surface.
At any instant the peripheral force acts on the mass at the contact point and accelerates it. The other mass would not be accerated. This would mean the center point of the dumb bell would accelerate at half the periphery.
You have to imagine at the next instant the masses are again normal.

So the force acts on half the rotation inertia to produce rotation and the center velocity will be half the peripheral speed.

TBC (must go)

blogfast25 - 10-3-2016 at 13:18

Quote: Originally posted by wg48

I don't think you can consider the force to act at the center.
But it does depend on what's you mean by that.

For the translational movement in the x-direction, simply consider the balance of forces in that direction to get the Newtonian equation of motion:

$$ma_x=\Sigma F_x$$

Here:

$$\Sigma F_x=F_F=-\mu_k mg$$

So:

$$a_x=-\mu_k g$$

So for the translational part it doesn't matter whether the friction force runs through the CoG or not.

F_F also exerts a moment about the horizontal axis going the CoG:

$$\tau=F_F R$$

That causes the deceleration of rotation:

$$I\alpha=\tau=F_F R$$

This causes the angular deceleration.

[Edited on 10-3-2016 by blogfast25]

wg48 - 11-3-2016 at 12:45

Thanks for your explanation Blogfest and woelen . I think what you mean is the peripheral force f can be resolved in to or equivalent to a force f acting at the c of g and a torque at the c of g of f x radius of the sphere. What I wanted to understand was what and how the peripheral. force can be resolved into.

My dumb bell model seems to have worked. I conclude from it that the peripheral force results in a rolling motion as if the sphere is rolling along an invisible surface on the opposite side of the sphere to the peripheral force. Though I have not fully checked it I think your translation and torque equivalence means the same as I am saying.

blogfast25 - 11-3-2016 at 15:53

Quote: Originally posted by wg48

I think what you mean is the peripheral force f can be resolved in to or equivalent to a force f acting at the c of g and a torque at the c of g of f x radius of the sphere. What I wanted to understand was what and how the peripheral. force can be resolved into.

I'm still not sure what you mean by 'resolving' here.

Take this classic balancing problem:

Assume the beam is massless and ignore all friction, air drag etc.

With no forces acting in the x-direction, there no acceleration or deceleration along the x-axis.

In the y-direction, we have:

$$\Sigma F_y=F-F_1-F_2=0$$

... in order to have no acceleration in the y-direction.

F₁ and F₂ also exert moments about the pivoting point, so that to avoid any rotational acceleration:

$$\tau_1-\tau_2=0$$

$$F_1L_1=F_2L_2$$

The forces F₁ and F₂ 'act' as forces and moments all at once. There's no 'resolving', even if they don't run through the CoG of the system.

[Edited on 11-3-2016 by blogfast25]

wg48 - 11-3-2016 at 17:58

Quote: Originally posted by blogfast25

Quote: Originally posted by wg48

I'm still not sure what you mean by 'resolving' here.

Take this classic balancing problem:

Assume the beam is massless and ignore all friction, air drag etc.

With no forces acting in the x-direction, there no acceleration or deceleration along the x-axis.

In the y-direction, we have:

$$\Sigma F_y=F-F_1-F_2=0$$

... in order to have no acceleration in the y-direction.

F₁ and F₂ also exert moments about the pivoting point, so that to avoid any rotational acceleration:

$$\tau_1-\tau_2=0$$

$$F_1L_1=F_2L_2$$

The forces F₁ and F₂ 'act' as forces and moments all at once. There's no 'resolving', even if they don't run through the CoG of the system.

[Edited on 11-3-2016 by blogfast25]

If the first sentence of my post when I used resolved does not give you my meaning I am almost lost on how else I could help.

I don's understand the significance of the beam example you gave. Yes no resolving of forces was required.

When you replaced the peripheral force with a force and a torque do you a have a term for that process.

blogfast25 - 11-3-2016 at 18:27

Quote: Originally posted by wg48

When you replaced the peripheral force with a force and a torque do you a have a term for that process?

No. Forces can take part in Newtonian equations of translation and Newtonian equations of rotation at the same time. There's no 'replacing': the friction force F_F provides a decelerating force for the translation and a decelerating torque for the rotation, at once. It's very common.

The significance of the beam example is that here we're looking at the forces and moments needed so that the state of motion (immobile) doesn't change. For that to happen, forces as well as moments have to be considered, both provided by the same forces F₁ and F₂.