Spherical ball rolling back and forth in a bowl

[zenterix]

Homework Statement

A spherical ball of radius $r$ and mass $M$, moving under the influence of gravity, rolls back and forth without slipping across the center of a bowl which is itself spherical with a larger radius $R$. The position of the ball can be described by the angle $\theta$ between the vertical and a line drawn from the center of curvature of the bowl to the center of mass of the ball. Show that, for small displacements, $\theta(t)$ is harmonic and find the frequency of oscillation.

Relevant Equations

The answer to this question is

$$\omega=\sqrt{\frac{5}{7}\frac{g}{R-r}}$$

Though I am interested in directly answering the question above, I am more interested initially in general calculations for this problem. Below I try to figure out what the normal force on the ball is based on the angle $\theta$. The calculation seems incorrect and I would like to know where I went wrong.

Here is what I have done.

$$\vec{N}(\theta)-mg\cos{\theta}\hat{r}=-R\dot{\theta}(t)\hat{r}\tag{1}$$

$$\implies \dot{\theta}(t)=\sqrt{\frac{mg\cos{\theta}-N(\theta)}{R}}\tag{2}$$

Since the spherical ball is rolling, I assume there is a friction force $f_S=\mu N(\theta)$ providing the associated torque.

$$mg\sin{\theta}(-\hat{\theta})+f_S\hat{\theta}=m\vec{a}=m(R-r)\ddot{\theta}(t)\hat{\theta}\tag{3}$$

$$\implies \ddot{\theta}(t)=\frac{_mg\sin{\theta}+\mu N(\theta)}{m(R-r)}\tag{4}$$

I then considered an angle $\gamma$ that measures the amount of rotation the sphere has undergone.

$$\vec{\tau}_{f_S}=I\vec{\alpha}\tag{5}$$

$$R\hat{r}\times f_S(\gamma)\hat{\theta}=Rf_S(\gamma)\hat{k}=I\ddot{\gamma}\hat{k}\tag{6}$$

The no slipping condition I came up with is

$$Rd\theta=rd\gamma\tag{7}$$

$$\implies R\dot{\theta}=r\dot{\gamma}\implies R\ddot{\theta}=r\ddot{\gamma}\tag{8}$$

At this point, I used (8) together with (4) and (6)

$$\frac{R\mu N(\theta)}{I}=\ddot{\gamma}=\frac{R}{r}\ddot{\theta}=\frac{R}{r}\frac{(-mg\sin{\theta}+\mu N(\theta)}{m(R-r)}\tag{9}$$

which leads to

$$N(\theta)=\frac{-Img\sin{\theta}}{\mu(rm(R-r)-I)}\tag{10}$$

$$=-\frac{2mrg\sin{\theta}}{\mu(5R-7r)}\tag{11}$$

which does not seem to be correct at all based on physical intuition (when $\theta=0$ we should have a the normal force at its highest).

 

[kuruman]

Please fix the LaTeX to increase legibility.

 

[haruspex]

Equation (1) is missing a power of 2 (corrected in (2)) and a mass on the RHS.
Gravity has a positive component in the $\hat r$ direction, not $-\hat r$. This leads to a sign error in (2). (When at the bottom of the bowl, is the normal force more or less than mg?)

 

[haruspex]

Since the spherical ball is rolling, I assume there is a friction force $f_S=\mu N(\theta)$ providing the associated torque.

Why? Is it on the point of slipping?

 

[TSny]

The no slipping condition I came up with is

$$Rd\theta=rd\gamma\tag{7}$$

This needs correction. Let $ds$ be a small displacement of the center of the ball while rolling without slipping in the bowl. How can you express $ds$ in terms of $d\gamma$? How can you express the same $ds$ in terms of $d \theta$?

 

[kuruman]

I think you need to reconsider equation (7). You are interested in the motion of the sphere's center, because eventually you are to liken it to the motion of a pendulum.

Note that when the angular displacement of the sphere's center is $\theta$, the center is displaced along an arc of length $s=(R-r)\theta$. So when he sphere rolls without slipping and rotates by angle $\gamma$ about its center, what is the arc $s'$ along which this center is displaced in terms of $\gamma$? Setting $s'=s$ should give you the appropriate constraint.

On edit: I see that @TSny preempted me (again) but I will let my post stand.

 

[zenterix]

Why? Is it on the point of slipping?

The problem states that the ball is rolling.

As far as I can tell, if there is no friction between the sphere and the bowl, then the sphere just slides without rolling.

 

[kuruman]

The problem states that the ball is rolling.

As far as I can tell, if there is no friction between the sphere and the bowl, then the sphere just slides without rolling.

The ball cannot roll without friction. So the friction is implicit.

 

[zenterix]

This needs correction. Let $ds$ be a small displacement of the center of the ball while rolling without slipping in the bowl. How can you express $ds$ in terms of $d\gamma$? How can you express the same $ds$ in terms of $d \theta$?

The linear displacement of the center of the ball is

$$ds_{center}=(R-r)d\theta$$

The length of the arc traversed by the ball is

$$ds_{ball}=rd\gamma$$

No slipping means $$(R-r)d\theta=rd\gamma$$

Previously I had used $dS=Rd\theta$, which is incorrect.

Making this change and redoing the calculations generates

$$N(\theta)=\frac{2mrg\sin{\theta}}{2r-5R}$$

which still does not seem correct.

 

[zenterix]

The ball cannot roll without friction. So the friction is implicit.

Yes, indeed, this is what I thought and is why I introduced friction into the calculations.

 

[kuruman]

which still does not seem correct.

Please show the details of your calculations. We cannot help you if you don't show what you did. Specifically, how did you introduce the corrections suggested by @haruspex in post #3? Also, when you write down torques, you need to be specific about what point you are calculating them.

 

[zenterix]

Here are the calculations once more

In post #3 Haruspex mentioned mistakes in equations (1) and (2). Here are the corrected versions

$$mg\cos{\theta}\hat{r}+N(\theta)\hat{r}=-mR\dot{\theta}^2(t)\tag{1}$$

$$\implies \dot{\theta}(t)=\sqrt{-\frac{mg\cos{\theta}+N(\theta)}{R}}\tag{2}$$

These equations are not, however, used in the subsequent calculations.

Using the 2nd law once more but in the tangential direction we have

$$mgsin{\theta}(-\hat{\theta})+f_S\hat{\theta}=m\vec{a}=m(R-r)\ddot{\theta}(t)\hat{\theta}$$

$$\implies \ddot{\theta}(t)=\frac{-mg\sin{\theta}+\mu N(\theta)}{m(R-r)}$$

where I have used $$f_S=\mu N(\theta)$$ as the friction force.

Torque at the point of contact with the bowl relative to the center of the ball is

$$\vec{\tau}=I\ddot{\gamma}\hat{\theta}$$

$$\vec{\tau}=r\hat{r}\times f_S\hat{\theta}=rf_S(\theta)\hat{k}=r\mu N(\theta)$$

Thus, equating the two equations above we get

$$\ddot{\gamma}(t)=\frac{r\mu N(\theta)}{I}$$

The no slipping condition is

$$(R-r)d\theta=rd\gamma$$

$$\implies (R-r)\dot{\theta}(t)=r\dot{\gamma}(t)$$

$$\implies (R-r)\ddot{\theta}(t)=r\ddot{\gamma}(t)$$

$$\implies \ddot{\gamma}(t)=\frac{R-r}{r}\ddot{\theta}(t)=\frac{R-r}{r}\left (\frac{-mg\sin{\theta}+\mu N(\theta)}{m(R-r)}\right )$$

We have two different expressions for $\ddot{\gamma}$ with a single unknown $N(\theta)$.

If we equate the two expressions then we have

$$\frac{R-r}{r}\frac{-mg\sin{\theta}+\mu N(\theta)}{m(R-r)}=\frac{r\mu N(\theta)}{I}$$

After solving for $N(\theta)$ we get

$$N(\theta)=\frac{Img\sin{\theta}}{I-r^2m}$$

and if we sub in the moment of inertia of the solid sphere, which is $I=\frac{2}{5}mr^2$, we get

$$N(\theta)=-\frac{2mg\sin{\theta}}{3\mu}$$

 

[zenterix]

I have fixed equations above regarding torque. That part should be correct now.

 

[kuruman]

I can see the mistake already in the calculations above: the part about the torque is incorrect. Let me try to correct it.

If you mean $~f_S=\mu N(\theta)$, yes. This equation is valid only when the sphere is on the verge of starting to slip as it rolls. There is no indication that this is the case here.

 

[zenterix]

@kuruman

I was referring to the fact that I said the torque was relative to the center of the ball's trajectory but it is actually relative to the center of the ball.

$$\tau=I\ddot{\gamma}\hat{\theta}$$

$$=r\hat{r}\times f_S(\theta)\hat{\theta}=rf_S(\theta)\hat{k}$$

$$\implies \ddot{\gamma}(t)=\frac{r\mu N(\theta)}{I}$$

As for the equation $f_S=\mu N(\theta)$ I am under the impression that the velocity of the point of contact relative to the ground (the bowl in this case) is zero and hence that the friction is always static.

Note that the point of contact is "trying" to slip: gravity is pulling the ball towards sliding down, and this is when static friction comes into play at the contact point.

As stated in the post above with the calculations, the current expression for the normal force as a function of $\theta$ is

$$N(\theta)=-\frac{2mg\sin{\theta}}{3}$$

which is simpler than before but the issue I have is the sine. If it were a cosine I would be happy. After all, the sine says that at the bottom of the trajectory the normal force is zero, and that for $\theta=\pi/2$ the normal force would be at a max. Should it not be the opposite?

 

[kuruman]

The instantaneous angular momentum of the ball is the sum of

• the angular momentum of the center of mass of the ball with respect to the center of the shell. This is sometimes given the name "orbital" angular momentum, $\mathbf {L}_{\text{orb}}.$
• the angular momentum about the center of mass of the ball. This is sometimes given the name "spin" angular momentum, $\mathbf {L}_{\text{spin}}.$

The torque that changes the orbital angular momentum (related to $\ddot {\theta}$) is generated by $mg$ and is referenced at the center of the shell. The torque that changes the spin angular momentum (related to $\ddot {\gamma}$) is generated by $f_s$ and is referenced at the center of the ball.

You cannot say that $f_s=\mu N$. The force of static friction is whatever is necessary to provide the observed acceleration. Perhaps it will be a good idea to answer the following question and see how $f_s$ works in a simpler case first:

A solid sphere of mass $m$ and radius $R$ rolls without slipping down an incline of fixed angle $\theta$ relative to the horizontal. Find the force of static friction $f_s$.

 

[zenterix]

Suppose a solid sphere of radius $R$ is initially at rest on an inclined plane as shown below

Gravity does not produce torque about the CM of the ball. Friction at the point of contact with the inclined plane does.

Torque

$$\vec{\tau}_{cm,f_S}=R\hat{j}\times f_S\hat{i}$$

$$=-Rf_S\hat{k}$$

$$=I_{cm}\ddot{\gamma}\hat{k}$$

$$\implies \frac{2}{5}MR^2\ddot{\gamma}=-Rf_S$$

$$\implies \ddot{\gamma}(t)=-\frac{5f_S}{2MR}$$

No Slipping

$$Rd\gamma=ds_{cm}$$

$$\implies v_{cm}(t)=R\dot{\gamma}$$

$$\implies a_{cm}(t)=R\ddot{\gamma}$$

2nd Law

$$Mg\cos{\beta}=N$$

$$Mg\sin{\beta}\hat{i}+f_S\hat{i}=M\vec{a}_{cm}=MR\ddot{\gamma}(t)\hat{i}$$

$$\implies Mg\sin{\beta}+f_S=MR\ddot{\gamma}(t)$$

Torque + 2nd Law

$$Mg\sin{\beta}+f_S=MR\left ( -\frac{5f_S}{2MR}\right )$$

$$\implies f_S=-\frac{2}{7}Mg\sin{\beta}$$

Note

The normal force counterbalances the component of gravity perpendicular to the incline plane.

$$\vec{N}=Mg\cos{\beta}(-\hat{j})$$

Friction was obtained by relating angular acceleration (obtained from torque generated by friction) with the linear acceleration of the CM (obtained from 2nd law).

Note that if we accept that $f_S=\mu N$ then what happens is that $\mu$ is obtained as a result of the other calculations. That is, given that we assumed no slipping, $\mu$ must have a specific value for this to be true.

$$\mu_{required}=\frac{f_S}{N}=\frac{2}{7}\frac{\sin{\beta}}{\cos{\beta}}$$

$\mu$ represents the maximum static friction coefficient. Thus we must have

$$\mu_{required}\leq \mu$$

Here is a plot of $\mu_{required}=\frac{2}{7}\frac{\sin{\beta}}{\cos{\beta}}$

The larger $\beta$ is, the larger the required $\mu$ for there to be no slipping.

If $\mu_{required}$ becomes larger than $\mu$ there will be slipping.

Alternatively, if $\mu$ is given, then we have

1) As before, from torque we obtain the angular acceleration of the sphere

$$\ddot{\gamma}(t)=-\frac{5f_S}{2MR}$$

2) From the 2nd law

$$N=-Mg\cos{\beta}$$

$$Mg\sin{\beta}+f_S=Ma_{cm}$$

$$\implies a_{cm}=\frac{Mg\sin{\beta}+f_S}{M}$$

3) At this point we have

(i) $N$ is a constant force.

(ii) $\ddot{\gamma}$ depends on $f_S$.

(iii) $a_{cm}$ depends on $f_S$.

4) If $f_S=\mu N$ then we can determine all the variables.

 

[kuruman]

Several mistakes. Please try to be careful.
1. You have drawn $\mathbf{f}_s$ up the incline which means that $\mathbf{f}_s=-f_s\mathbf{\hat i}$ not what you have.
2. You have resolved the weight incorrectly. Check your trigonometry.

If μ is not this value then there will be slipping or skidding.

This is not true. What you want to say is that
As long as $f_s < \mu N$ there will be rolling without slipping and we don't care what the actual value of $\mu$ is and it should not enter the picture.

I have to sign off for now. See if you can apply what you learned to the original problem.

 

[zenterix]

I drew $f_S$ up the incline because that is what physically happens.

Mathematically such a vector can be represented as $f_S\hat{i}$ with $f_S<0$. This is exactly what I am doing.

Indeed I have exchanged cosine with sine and vice-versa in decomposing the gravitational force. I have fixed this in post #17.

It is true that $\mu$ represents a maximum static friction coefficient.

Since we have $f_S$ and $N$, the "required" static friction coefficient is obtained as I said

$$\mu_{required}=\frac{f_S}{N}=\frac{2}{7}\frac{\sin{\beta}}{\cos{\beta}}$$

and this must be compared to $\mu$, the maximum static friction coefficient. As long as $\mu_{required}\leq \mu$ we have no slipping.

 

[sophiecentaur]

Since the spherical ball is rolling, I assume there is a friction force

Imo, you can forget about the value of the 'friction force' as long as there is no slipping (specified in OP). So why would you expect to have μ turning up in almost every line of your workings? All you need to assume is that the tangential force on the ball is equal to the force causing the torque plus the force causing the 'linear' acceleration down the slope. That would be the same as for ideal matching concave and convex gears; linear velocity is the same on both contacting sides.
What happens when there's slippage is not part of the original question so why go there?

 

[kuruman]

The no slipping condition is usually written as $~f_s\leq \mu_sN.~$ In the case of the rolling ball within the shell, we are told that the ball rolls without slipping, therefore you may safely assume that the condition is satisfied throughout the motion.

Now, can you apply all this to the problem at hand? Note that, as in the case of the exercise you solved with the ball rolling down the incline, the coefficient of static friction $\mu_s$ is not and should not be part of the picture.

Also see @sophiecentaur's post #20.

 

[zenterix]

Imo, you can forget about the value of the 'friction force' as long as there is no slipping (specified in OP). So why would you expect to have μ turning up in almost every line of your workings? All you need to assume is that the tangential force on the ball is equal to the force causing the torque plus the force causing the 'linear' acceleration down the slope. That would be the same as for ideal matching concave and convex gears; linear velocity is the same on both contacting sides.
What happens when there's slippage is not part of the original question so why go there?

The original question was the starting point for me to ask questions of my own such as what is the normal force and what is the friction force.

The original problem seems to be relatively simpler. Here is how I tried to solve it

Consider the sphere when it reaches the largest amplitude and is at rest before rolling back down. Take this to be $t=0$. At this instant, the sphere makes an angle $\theta_0$ with a vertical line, there is no kinetic energy, and the potential energy is

$$U(\theta_0)=mg(R-r)(1-\cos{(\theta_0)}$$

Now consider a time $t>0$. The total work done on the system is

$$W=\frac{mv_{S,cm}^2}{2}+\frac{I_{cm}\omega_{cm}^2}{2}$$

where $S$ denotes the center of the circle that the trajectory of the sphere is on. That is, the total work is composed of a translational term relative to $S$ and a rotational term relative to the CM of the sphere.

$$W=\frac{m(R-r)^2\dot{\theta}^2}{2}+\frac{mr^2}{5}\dot{\gamma}^2$$

At this point we use the no-slipping condition $\dot{\gamma}=\frac{R-r}{r}\dot{\theta}$.

$$W=\frac{m(R-r)^2\dot{\theta}^2}{2}+\frac{mr^2}{5}\frac{(R-r)^2}{r^2}\dot{\theta}^2$$

$$=\frac{7}{10}\dot{\theta}^2m(R-r)^2$$

We now equate this to $-\Delta U$, the change in potential energy, which is

$$-\Delta U= U_i-U_f=mg(R-r)(1-\cos{\theta_0})-mg(R-r)(1-\cos{\theta})$$

Using the approximation that for small $\theta$ we have

$$\cos{\theta}\approx 1-\frac{\theta^2}{2}$$

we get

$$-\Delta U=mg(R-r)\left (\frac{\theta_0^2}{2}-\frac{\theta^2}{2}\right )$$

And so

$$\Delta K=-\Delta U$$

$$\implies \frac{7}{10}\dot{\theta}^2m(R-r)^2=mg(R-r)\left (\frac{\theta_0^2}{2}-\frac{\theta^2}{2}\right )$$

$$\dot{\theta}=\sqrt{\frac{5}{7}\frac{g}{R-r}(\theta_0^2-\theta^2)}$$

Now, this doesn't seem correct. I expected to have gotten an $\ddot{\theta}$ on the left-hand side.

 

[kuruman]

You are using the small angle approximation in the wrong place. Remember that in simple pendulum you approximate the equation of motion (EOM) for small angles, not the change in potential energy: $$\ddot {\theta}+\frac{g}{L}\sin\theta\rightarrow \ddot {\theta}+\frac{g}{L}\theta.$$Here you need to do the same. First find the EOM and then see what happens at small angles.

Here is an alternate method for deriving the EOM that relies on energy considerations and bypasses FBDs.

1. Write an expression for energy conservation from the starting point to some angle $\theta.$
2. Define $\dot{\theta}\equiv \omega$ and find an expression for $\omega^2$.
3. Use the chain rule to find a relation between $\ddot {\theta}$ and $\omega^2$.

 

[zenterix]

Consider total mechanical energy when the angle is $\theta$.

$$E_m=\frac{7}{10}m(R-r)^2\dot{\theta}^2+mg(R-r)(1-\cos{\theta})$$

The first term on the right-hand side was computed in post #22 (it is the sum of translational kinetic energy relative to the center of the circle the sphere is moving along, plus rotational kinetic energy about the CM), and the second term is gravitational potential energy.

There are no non-conservative forces doing work (note that although there is friction creating torque, it is static friction, which does no work) so mechanical energy is conserved.

$$\frac{dE_m}{dt}=0\implies \frac{7}{5}m(R-r)^2\ddot{\theta}\dot{\theta}+mg(R-r)\sin{\theta}\dot{\theta}=0$$

At this point we use the small angle approximation $\sin{\theta}\approx \theta$ to obtain

$$\implies \ddot{\theta}+\frac{5}{7}\frac{g}{R-r}\theta=0$$

which we know describes simple harmonic motion with angular frequency

$$\omega=\sqrt{\frac{5}{7}\frac{g}{R-r}}$$

 

[kuruman]

That works.