Homework Help: Frictionless mass sliding in rotating pipe

1. Jun 14, 2017

Feodalherren

1. The problem statement, all variables and given/known data

A pipe is spinning around with angular velocity ω = 2 rad/s about its fixed end. Inside the pipe is a frictionless block that is free to slide. The block is located 1 m from the fixed end of the pipe when the angle from the vertical, θ = π/2 (horizontal). What is the radial acceleration of the block at this instant?

2. Relevant equations

3. The attempt at a solution

My immediate intuition is that there is no acceleration along the pipe, but how do I show that?

I tried to draw a FBD but I can't even come up with any forces to write on it. There is no centripetal force, there is no friction... there is nothing.

It's sort of like 2.33 on here
https://www.slideshare.net/anshukg/an-introduction-to-mechanics-daniel-kleppner-solutions-part1

But I'm not really able to follow the solution there.

Last edited: Jun 14, 2017
2. Jun 14, 2017

haruspex

Mine too.
This seems like a reasonable way:
... except that you should also show there is no tangential acceleration.

3. Jun 14, 2017

Feodalherren

So how do I show that there is no tangential acceleration along the pipe?

4. Jun 14, 2017

haruspex

Tangential is not along the pipe. What do you know about the pipe's movements?

5. Jun 14, 2017

Feodalherren

Ops I misspoke. I only know that it's rotating at a constant speed.

6. Jun 14, 2017

haruspex

So what is happening to the vertical velocity?

7. Jun 14, 2017

Feodalherren

Slowing down. I just can't connect the dots with how that does anything to show me that there is no acceleration radially for the mass.

8. Jun 14, 2017

haruspex

It is? Why?
The questions just says acceleration. It does not restrict to radial acceleration.

9. Jun 14, 2017

Feodalherren

Because at the instant when it is perpendicular to the vertical all the velocity is in the Y direction, as it passes that position some velocity will be in x and some in y and since the angular speed is constant that means that it must accelerate in x and decelerate in y.

Ops that was a mistake from my side. I translated the problem from German and missed that part. It specifically asks to solve for the acceleration along the pipe i.e. radial.

10. Jun 14, 2017

Feodalherren

I suppose what I descrived above is some dt after the case that I am studying, so in this case there is no tangential acceleration. I just fail to see how I'm supposed to "show" any of this mathematically. It ends up being an empty free body diagram.

11. Jun 14, 2017

haruspex

That is a bit ambiguous. It could mean the component of acceleration in the radial direction, or it could mean $\ddot r$. Is it clearer in the German?
Yes.
As you have - there are no radial forces, so no radial acceleration (unless the other interpretation is intended).

12. Jun 14, 2017

Feodalherren

Yes, it literatelly says $\ddot r$ I just don't know how to use LATEX.

So just draw up an FBD, write something like "no forces in radial direction, therefore radial acceleration = 0."? Almost seems too easy :D.

13. Jun 14, 2017

haruspex

$\ddot r$ is not the same as radial component of acceleration, so the answer is not zero.
E.g. if $\ddot r$ were continuously zero then r would not change, so there would be centripetal acceleration.

14. Jun 14, 2017

Feodalherren

Ah yeah, you're right of course. Well then I'm completely lost.

15. Jun 14, 2017

haruspex

No need to be lost. My example in the previous post was a bit of a hint.
If $\ddot r$ were zero, what would radial acceleration be?
So if the radial acceleration is zero....?

16. Jun 14, 2017

Feodalherren

Hmm.. To radial is zero, angular acceleration is zero since we have a constant omega. I just can't see it. With a pendulum we'd have centripetal acceleration but in this case I don't know. Since omega is constant I assume tangential acceleration is zero too.

17. Jun 14, 2017

haruspex

Take a look at the expressions for radial and tangential acceleration in polar coordinates at https://en.m.wikipedia.org/wiki/Polar_coordinate_system#Centrifugal_and_Coriolis_terms.

18. Jun 15, 2017

Feodalherren

Ok so from an inertial reference frame Newton's second law in polar coordinates becomes

F=m[ (r''- rθ'^2)r + (rθ'' + 2r'θ')θ]

therefore

a = (r''- rθ'^2)r + (rθ'' + 2r'θ')θ

We know that the radial component of the acceleration is zero from the FBD thus:

r'' - rθ'^2 = 0

then r'' = rθ'^2 = (1m)(2rad/s)^2 = 4m/s^2

is this the final answer though, since the question was asking for r'', I assume I do not need to solve for the θ term of a?

What exactly does r'' mean in physical terms? Isn't r'' the radial acceleration, and we just looked at the FBD and determined that it's zero!? What is the difference between the r component of a and r''?

Last edited: Jun 15, 2017
19. Jun 15, 2017

haruspex

Yes.
$\dot r$ is the rate of change of r, and $\ddot r$ is the rate of change of $\dot r$. Not sure that I can give any better physical interpretation.
If you take a non-inertial frame of reference, Cartesian coordinates centred on the mass and oriented with x along the pipe, the mass appears to be thrown in the x direction by a centrifugal force. The resulting acceleration is $\ddot r$.

You could also consider throwing away the pipe. A mass travels at constant speed in a straight line past you, missing you by some distance. If r(t) is the distance from you to the mass it clearly reaches a minimum before increasing again. $\ddot r$ is always positive.

20. Jun 15, 2017

Feodalherren

Thank you! I'm still not sure that I understand the physical meaning of it though - I understand the derivative notation. What I don't understand is that when we set up the problem we said that the radial acceleration is zero at the point in time when we're looking at the problem - that far I am with you.

When we started setting up the problem we got the equation a = (r''- rθ'^2)r + (rθ'' + 2r'θ')θ, where r is the radial component in polar coordinates. So we know that radial acceleration is zero and thus we were able to set up the problem and solve it. But what is the difference between radial acceleration and r'', it seems to me that r'' IS the radial acceleration?

21. Jun 15, 2017

kuruman

Perhaps you will be able to understand the two terms of the radial acceleration better if you imagine what the two terms in it are individually responsible for. It can be rewritten as $a_r = \ddot{r}-\omega^2~r$.

Imagine dropping a ball from rest near the surface of the Earth. In spherical coordinates there is only a radial component of the acceleration (neglecting Coriolis effects) , given by $\ddot{r} = -g$. This term is responsible for changing the magnitude of the radial component of the velocity.

Now imagine an object going around in a circle of radius $r$. In this case $\ddot{r}$ is zero and all you have is the centripetal acceleration described by the second term. It is responsible for changing the direction of the tangential component of the velocity.

It might be a good exercise for you to derive an expression in spherical coordinates for the acceleration of a projectile near the surface of the Earth when it is at position $\vec{r}$ from the origin moving with velocity $\vec{v}$.

On edit: Can you identify the tangential and Coriolis terms in $a_{\theta} = r\ddot{\theta} +2~\dot{r}~\dot{\theta}$? Which one changes the magnitude of the tangential component of the velocity and which one changes the direction of the radial component of the velocity?

Last edited: Jun 15, 2017
22. Jun 15, 2017

jbriggs444

"radial acceleration" is the component of acceleration in the direction that is [momentarily] radial.

Suppose that you have a front row seat near the center line of a hockey rink. You watch as the puck moves left to right in front of you from one goal toward the opposite goal. The distance from the puck to your seat will decrease, first rapidly then slowly as the puck approaches the center line in front of you. After it crosses the center line, the distance from the puck to your seat will increase, first slowly then rapidly.

The second derivative of radial distance (the distance from your seat to the puck) with respect to time has been positive throughout. But that is not its "radial acceleration". That is the second derivative of its radial distance.

The puck has had zero acceleration throughout this exercise. The fact that we might choose to use polar coordinates to describe its position does not change this. Its acceleration is still zero. The fact that the individual polar coordinates describing that position do not change linearly over time does not alter this.

At the moment the puck crosses the center line, the radial direction is directly along the center line. The puck's acceleration is the zero vector. The projection of the zero vector in the radial direction is... zero.

So the puck's "radial acceleration" is zero at that moment. (And at every other moment during this exercise).

23. Jun 15, 2017

Feodalherren

Aaah I get it now, it's an issue of reference frames. Very good explanation, thanks!

I think $2~\dot{r}~\dot{\theta}$ is the component that changes the magnitude, I'm not really sure though.