Dynamics of Circular Motion

[JuanMa2409]

Homework Statement

A very small ball of mass 𝑚 is threaded on a fixed vertical ring of radius
𝑅, along which it can slide without friction. The ball starts at rest from the highest point of the ring (position 𝐴, corresponding to 𝜃=0), and begins to slide.

Question:
At which point(s) along the trajectory is the ball subject to a purely horizontal acceleration?

Relevant Equations

Circular motion equations, Newton's Laws, Work-Energy theorem

To solve this problem first I try to use polar coordinates, then I write the forces that I obtain in the free body diagram. That are the gravitational force and the tension force. With this using the second Newton's law I write the forces that are equal to the acceleration in polar coordinates times the mass. Next, I write the Work-Energy theorem to obtain the tangential velocity, and with that result I replace it in the equation given in the previous part. But the problem it's that I don't understand what condition I need to satisfy to get only a horizontal component of acceleration. Maybe I think that I don't get at all the concept of acceleration in a circular movement, anyways what I understand is that the vector of acceleration is equal to the sum of tangential acceleration and centripetal acceleration.
Please, I would appreciate if someone could guide me in this part, or correct me if my reasoning is incorrect.

 

[PeroK]

If you want to calculate the horizontal and vertical components of acceleration, then polar coordinates are not very useful.

Try the same method using Cartesian coordinates. The acceleration is horizontal when the vertical component is zero.

 

[kuruman]

But the problem it's that I don't understand what condition I need to satisfy to get only a horizontal component of acceleration.

The same condition that is satisfied when a book is at rest on a table top. Here you have a bead and a ring but the condition is the same.

 

[JuanMa2409]

If you want to calculate the horizontal and vertical components of acceleration, then polar coordinates are not very useful.

Try the same method using Cartesian coordinates. The acceleration is horizontal when the vertical component is zero.

Thanks for your tip, I tried that method and I don't get any result yet.
Here is my attempt:

Where:
- a_x is acceleration in x.
- a_x is acceleration in y.
- T is tension.

I know that I need to calculate the points where the acceleration is purely horizontal, so I think that I'm searching for an angle. But even when I found the equations, I think that I have more unknowns than equations. I also try to use the first equation knowing that T = -ma_x/senΘ, and then replace it into the second equation, but I only get a new equation with more unknowns as the previous one.
Maybe I'm mistaken about the forces or the theorem; I'm not sure. I'd appreciate your guidance.

 

[PeroK]

Thanks for your tip, I tried that method and I don't get any result yet.
Here is my attempt:
View attachment 363552
Where:
- a_x is acceleration in x.
- a_x is acceleration in y.
- T is tension.

I know that I need to calculate the points where the acceleration is purely horizontal, so I think that I'm searching for an angle. But even when I found the equations, I think that I have more unknowns than equations. I also try to use the first equation knowing that T = -ma_x/senΘ, and then replace it into the second equation, but I only get a new equation with more unknowns as the previous one.
Maybe I'm mistaken about the forces or the theorem; I'm not sure. I'd appreciate your guidance.

I didn't think of doing it your way. Let me have a look at that method. The approach I took was:

You have correctly determined the speed of the ball as a function of the angle:
$$v^2 = 2gR(1- \cos \theta)$$I would tackle the problem using kinematics. If we take the origin as the centre of the ring, we have:
$$y = R\cos \theta$$$$\dot y = -(R\sin \theta) \dot \theta$$Can you calculate $\ddot y$? Be careful to use the chain rule properly:
$$\ddot y = ?$$

 

[PeroK]

I think that looking at forces adds unnecessary calculations, as you know the acceleration of the ball. There is no need to calculate the constraining force $T$. Instead, an alternative to my method is simply to add the centripetal and tangential accelerations.

You know the centripetal acceleration of the ball:
$$a_c = \frac{v^2}{R}$$And you know the tangential acceleration of the ball:
$$a_t = g|\sin \theta|$$The tricky part is the trigonometry, because $\theta$ is not the usual polar angle. So, it's a little bit tricky to figure out the y-component of these accelerations.

Can you see approximately where the acceleration will be horizontal? Without doing specific calculations?

PS this is simpler than my first approach. This is the way I would do it. Just add the known vector accelerations.

 

[JuanMa2409]

I think that looking at forces adds unnecessary calculations, as you know the acceleration of the ball. There is no need to calculate the constraining force T. Instead, an alternative to my method is simply to add the centripetal and tangential accelerations.

You know the centripetal acceleration of the ball:
ac=v2RAnd you know the tangential acceleration of the ball:
at=g|sin⁡θ|The tricky part is the trigonometry, because θ is not the usual polar angle. So, it's a little bit tricky to figure out the y-component of these accelerations.

Can you see approximately where the acceleration will be horizontal? Without doing specific calculations?

PS this is simpler than my first approach. This is the way I would do it. Just add the known vector accelerations.

You're right, that's a easier way.
So the acceleration in $y$ is:
$$\ddot y=−(R\cos \theta)\dot \theta$$
And $\dot \theta$ is the angular speed $\omega$, so:
$$\ddot y=−(R\cos⁡ \theta)\omega$$
$$=−\frac{R\cos⁡ \theta)v^2}{R^2}$$
$$=−\frac{v^2\cos⁡ \theta}{R}$$
Knowing the velocity of the ball, I have that:
$$\ddot y=−\frac{2gR(1−\cos⁡ \theta)\cos \theta⁡}{R}$$
$$=−2g(\cos⁡ \theta−\cos^2⁡ \theta)$$
Then the if $\ddot y = 0$:
$$−2g(\cos⁡ \theta−\cos^2⁡ \theta)=0$$
$$\cos⁡ \theta(1−\cos⁡ \theta)=0$$
$$1−\cos⁡ \theta=0$$
$$\cos⁡ \theta=1$$
Then $\theta=0$.
But it has no sense, because the position of the ball in $\theta = 0$ is on the top, and if it is on the top then the centripetal acceleration points down. Answering your question about see approximately where the acceleration will be horizontal, I don't see any point where this condition is satisfied. Because for example if the ball is on the left or on the right the tangential acceleration points up or down even if the centripetal is only horizontal. Therefore, I think perhaps the problem has no solution.
Anyways if I'm doing a mistake in my reasoning, let me now.
And I'll try your other method by adding the accelerations, the only thing that I don't understand is which equation do I need to equal to zero, due to if I say that $a_c+a_t=0$ I'm supposing that the entire acceleration is equal to zero, right?

 

[berkeman]

I could be wrong, but at first glance to me it seems more like an intuitive problem rather than a quantitative one. The acceleration of gravity is pointing downwards, and that is modified by the constraints of motion along the ring. It seems to me that the points along the ring where the acceleration could be purely horizontal correspond to the acceleration of gravity there being...

 

[JuanMa2409]

I could be wrong, but at first glance to me it seems more like an intuitive problem rather than a quantitative one. The acceleration of gravity is pointing downwards, and that is modified by the constraints of motion along the ring. It seems to me that the points along the ring where the acceleration could be purely horizontal correspond to the acceleration of gravity there being...

Maybe you're right, but I'd like to also prove it.

 

[kuruman]

Maybe you're right, but I'd like to also prove it.

Here are the facts

• There are two forces acting on the mass, the weight $\mathbf W$ and the radial normal force $\mathbf N.$
• The acceleration is $\mathbf a=\dfrac{1}{m}(\mathbf W +\mathbf N).$
• For the acceleration to have only a horizontal component, it must be true that $0=a_y=\dfrac{1}{m}(W_y + N_y).$

All you need to do is find $W_y$ (trivial) and $N_y$ - not that tricky because you have already established that $v^2 /R= 2g(1- \cos \theta).$ Draw a FBD, find the magnitude $N$ and then find the vertical component $N_y$.

 

[PeroK]

You're right, that's a easier way.
So the acceleration in $y$ is:
$$\ddot y=−(R\cos \theta)\dot \theta$$

That equation is not correct. Note that it's not necessary to find the normal force. You already know the centripetal and tangential accelerations. You don't need to calculate the normal restraining force. Instead:

The tangential acceleration depends on the component of gravity tangential to the ring:
$$a_t = g\sin \theta$$The y-component of this is:$$a_t(y) = (g\sin\theta)(-\sin \theta) = -g\sin^2 \theta$$The y-component of the centripetal acceleration is:$$a_c(y) = (\frac{v^2}{R})(-\cos \theta) = -2g(1-\cos \theta)\cos \theta$$Now you just solve for $a_t(y) + a_c(y) = 0$.

To help you a bit more, let's look for points where the acceleration is entirely vertical. The x-component of the acceleration is:
$$a_x = a_t(x) + a_c(x) = (g\sin\theta)(\cos \theta) - 2g(1-\cos \theta)\sin \theta$$$$a_x = g\sin \theta(3\cos \theta - 2)$$The acceleration is purely vertical when $\sin \theta = 0$ or $\cos \theta = \frac 2 3$.

$sin \theta = 0$ corresponds to the points at the top and bottom of the ring - this is a degenerate solution where the acceleration is zero, so neither horizontal nor vertical. When $\cos \theta = \frac 2 3$, the ball has moved a height $\frac R 3$ below the top. At this point, the tangential acceleration is to the right and the centripetal acceleration is to the left and they have equal magnitude. By symmetry, there is a second point on the left, as the ball is moving back up to the top, where the acceleration is instantaneously vertical.

Going back to the vertical component, we can see that in the top half of the ring, both the tangential and centripetal acceleration point downwards. But, on the bottom half of the ring, the centripetal acceleration points upwards. At the bottom of the ring, the acceration is upwards, so there must be a point on each side where the vertical component of the acceleration changes sign and hence is instantaneously zero.

One of the first things I did with this problem is work out that the answer should be two points on the lower half of the ring. I can see immediately that your answer of $\cos \theta = 1$ is not right. Although, at the top of the ring, the acceleration is instantaneously zero, so that is a degenerate solution, as the horizontal acceleration is also zero.

 

[PeroK]

All you need to do is find $W_y$ (trivial) and $N_y$ - not that tricky because you have already established that $v^2 /R= 2g(1- \cos \theta).$ Draw a FBD, find the magnitude $N$ and then find the vertical component $N_y$.

There's definitely no need to calculate the normal constraining force. We already know the acceleration.

 

[kuruman]

There's definitely no need to calculate the normal constraining force. We already know the acceleration.

My response to that is that there is no need to

work out that the answer should be two points on the lower half of the ring.

The method outlined in post #10 leads to a quadratic equation in $\cos\theta$. One solution is $\theta = 0$, which is expected, the other is on the lower half of the ring.

I butted in only because the OP did not seem to appreciate the significance of

Can you see approximately where the acceleration will be horizontal? Without doing specific calculations?

I will now butt out on the principle that "too many cooks spoil the broth."