Man walking on a rotating merry-go-round

[JD_PM]

Homework Statement

I am studying the following exercise from Gregory's CM book (chapter 17th, Example 17.2).

Relevant Equations

Second law in a general non-inertial frame

Some information

Newton's second law in a non-inertial frame is given by:

Where:

1) $A$ is the translational acceleration, $\Omega$ the angular velocity and $\dot \Omega$ the angular acceleration (all relative to the inertial frame attached to the ground $F$).

2) r', v' and a' are the position, velocity and acceleration vectors, all relative to the frame attached to the roundabout $F'$ (and thus ' has nothing to do with derivatives on above equation).

I am studying the following exercise from Gregory's CM book (chapter 17th, Example 17.2).

Exercise statement:

OK so I get that:

$$m \vec a = -mg \vec e_3 + \vec X$$

Where $-mg \vec e_3$ is the gravitational force the Earth exerts on the man and $\vec X$ is the normal force the roundabout exerts on the man.

I mathematically understand how we get the final equation for the normal force $\vec X$, but I now want to understand the physics behind it.

I understand that we expect to get a term due to the gravitational force, another to the centrifugal force and another to the Coriolis Force.

But why are the last two negative? Why couldn't they be positive?

 

[PeroK]

Homework Statement:: I am studying the following exercise from Gregory's CM book (chapter 17th, Example 17.2).
Homework Equations:: Second law in a general non-inertial frame

Some information

Newton's second law in a non-inertial frame is given by:

View attachment 254244

Where:

1) $A$ is the translational acceleration, $\Omega$ the angular velocity and $\dot \Omega$ the angular acceleration (all relative to the inertial frame attached to the ground $F$).

2) r', v' and a' are the position, velocity and acceleration vectors, all relative to the inertial frame attached to the roundabout $F'$ (and thus ' has nothing to do with derivatives on above equation).

I am studying the following exercise from Gregory's CM book (chapter 17th, Example 17.2).

Exercise statement:

View attachment 254241
View attachment 254242
View attachment 254243

OK so I get that:

$$m \vec a = -mg \vec e_3 + \vec X$$

Where $-mg \vec e_3$ is the gravitational force the Earth exerts on the man and $\vec X$ is the normal force the roundabout exerts on the man.

I mathematically understand how we get the final equation for the normal force $\vec X$, but I now want to understand the physics behind it.

I understand that we expect to get a term due to the gravitational force, another to the centrifugal force and another to the Coriolis Force.

But why are the last two negative? Why couldn't they be positive?

When you ask why they are negative you really mean why is there a minus sign.

Plus or minus signs appear in equations often depending on how quantities are defined with respect to each other.

 

[TSny]

I mathematically understand how we get the final equation for the normal force $\vec X$, but I now want to understand the physics behind it.

I understand that we expect to get a term due to the gravitational force, another to the centrifugal force and another to the Coriolis Force.

But why are the last two negative? Why couldn't they be positive?

$\mathbf X$ is the true force that the roundabout exerts on the man. Relative to the inertial frame, part of the motion of the man is circular motion. So, there must be a centripetal component of force acting on the man. What direction is this in terms of the unit vectors $\mathbf {\hat e'_1}$, $\mathbf {\hat e'_2}$ , and $\mathbf {\hat e_3}$?

Also, from the point of view of the inertial frame, the speed of the man due to rotation decreases as he walks toward the center (from point a to point b in the figure below).

So, there is a component of acceleration opposite to the arrows shown in the figure. What direction is this in terms of the unit vectors $\mathbf {\hat e'_1}$, $\mathbf {\hat e'_2}$ , and $\mathbf {\hat e_3}$?

 

[JD_PM]

$\mathbf X$ is the true force that the roundabout exerts on the man.

When you said true you meant net, isn't it?

Relative to the inertial frame, part of the motion of the man is circular motion. So, there must be a centripetal component of force acting on the man. What direction is this in terms of the unit vectors $\mathbf {\hat e'_1}$, $\mathbf {\hat e'_2}$ , and $\mathbf {\hat e_3}$?

I expect the centripetal component of force acting on the man (wrt the inertial frame) to have the form:

$$\vec X_c = -A \ \mathbf {\hat e_1}$$

And I expect the centripetal component of force acting on the man (wrt the non-inertial frame) to have the form:

$$\vec X_c = -A \ \mathbf {\hat e'_1}$$

I've taken outwards direction as positive. The man is dragged inwards.

Also, from the point of view of the inertial frame, the speed of the man due to rotation decreases as he walks toward the center (from point a to point b in the figure below).

View attachment 254255

So, there is a component of acceleration opposite to the arrows shown in the figure. What direction is this in terms of the unit vectors $\mathbf {\hat e'_1}$, $\mathbf {\hat e'_2}$ , and $\mathbf {\hat e_3}$?

Thanks for the diagram.

Here what you are asking (if I am not mistaken) is for the direction of the tangential acceleration.

Thus, I expect the tangential component of force acting on the man (wrt the non-inertial frame) to have the form:

$$\vec X_t = - B \ \mathbf {\hat e'_2} $$

Mmm so with this exercise you tried to show me that the -ive sign in front of the $\mathbf {\hat e'_1}$ is due to the centripetal acceleration dragging the man inwards and that the -ive sign in front of the $\mathbf {\hat e'_2}$ is due to the tangential acceleration dragging the man to his left, didn't you?

Note $A$ and $B$ are constants.

 

[JD_PM]

When you ask why they are negative you really mean why is there a minus sign.

That's right

 

[PeroK]

That's right

It the quadratic formual, why is it $b^2 - 4ac$? Instead of ##b^2 + 4ac"?

 

[JD_PM]

It the quadratic formual, why is it $b^2 - 4ac$? Instead of ##b^2 + 4ac"?

But there's a purely mathematical reasoning for that. Physics is not needed to answer that question.

 

[PeroK]

But there's a purely mathematical reasoning for that. Physics is not needed to answer that question.

If you take the equation you posted in the first post:

That can be rewritten as:

$m[A - r' \times \dot{\Omega} - 2v'\times \Omega - (\Omega \times r') \times \Omega + a'] = F$

Now you have negative signs instead of positive signs.

 

[JD_PM]

If you take the equation you posted in the first post:

View attachment 254311

That can be rewritten as:

$m[A - r' \times \dot{\Omega} - 2v'\times \Omega - (\Omega \times r') \times \Omega + a'] = F$

Now you have negative signs instead of positive signs.

I see what you mean; if you change the signs of every element you end up with the same equation. For instance, the following equations are equivalent:

$$A + B - C + D = 0$$

$$-A - B + C - D = 0$$

But what I want is to interpret these signs; (for instance) why

$$A + B - C + D = 0$$

And not:

$$A + B - C - D = 0$$

There could be situations in which it is physically meaningless, but in the problem I posted it isn't.

 

[PeroK]

I see what you mean; if you change the signs of every element you end up with the same equation. For instance, the following equations are equivalent:

$$A + B - C + D = 0$$

$$-A - B + C - D = 0$$

But what I want is to interpret these signs; (for instance) why

$$A + B - C + D = 0$$

And not:

$$A + B - C - D = 0$$

There could be situations in which it is physically meaningless, but in the problem I posted it isn't.

The last term, the centripetal force is towards the centre, but the $e_1$ vector points away from the centre. So, that term is negative.

The middle term, the Coriolis force, is $\pm$ depending on the convention for $\Omega$. That said, what is physically meaningful is the direction of the Coriolis force relative to the directions of rotation and the man's velocity. If the man turns round and walks outward, then the Coriolis force changes direction.

That is worth thinking about.

Assuming here by "Coriolis force" we mean the real force on the man by the floor.

 

[JD_PM]

The last term, the centripetal force is towards the centre, but the $e_1$ vector points away from the centre. So, that term is negative.

I agree (by the way, didn't you mean $\mathbf {\hat e'_1}$?)

The middle term, the Coriolis force, is $\pm$ depending on the convention for $\Omega$.

Let us take counterclockwise direction as positive. The Coriolis force points to the right of the motion of the object then (https://en.wikipedia.org/wiki/Coriolis_force)

 

[PeroK]

I agree (by the way, didn't you mean $\mathbf {\hat e'_1}$?)
Let us take counterclockwise direction as positive. The Coriolis force points to the right of the motion of the object then (https://en.wikipedia.org/wiki/Coriolis_force)

Yes, the primed vector.

With the centrifugal/centripetal force we have a nice convention that one is the fictitious force and one is the real force. With the Coriolis force, we have only the convention that this is the fictitious force. I would tend to use "real" Coriolis force to identify the real force needed to keep the man on a straight path.

The thing to note is that as the man moves in his AM is decreasing $L = m\Omega r^2$. Without any lateral forces, he should be spinning faster than the merry-go-round. Therefore, he needs to push in the direction of rotation (i.e to his right) , and the the merry-go-round must react in the direction against the rotation (i.e. to his left). That's the real force, whatever you choose to call it.

If he walks outwards, the opposite is true as his AM is increasing.

 

[TSny]

When you said true you meant net, isn't it?

By "true" I mean "actual" as opposed to "fictitious" (or "apparent").

The force $\vec X$ represents the actual (true) force that the roundabout exerts on the person. The force of gravity is another "true force" acing on the person. $\vec X$ includes the normal force as well as the force of friction acting on the person. The friction force has both a centripetal component (in the $-\hat e_1$ direction) and a tangential component (in the $-\hat e_2$ direction).

Let's go back to your original question in post #1:

I mathematically understand how we get the final equation for the normal force $\vec X$, but I now want to understand the physics behind it.

I understand that we expect to get a term due to the gravitational force, another to the centrifugal force and another to the Coriolis Force.

But why are the last two negative? Why couldn't they be positive?

The terms "Coriolis" and "centrifugal" do not apply to $\vec X$ because $\vec X$ is a true force while the Coriolis force and the centrifugal force are fictitious forces that are present only in the rotating frame. The true force $\vec X$ has a normal component: $+mg \hat e_3$, a centripetal component: $-m\Omega^2 x_1 \hat e_1$ , and a tangential component: $-2m \Omega V \hat e_2$. The vertical component is the normal force while the other two components represent the friction force acting on the man. The $\hat e_1$ component has a negative sign because this component of friction provides the necessary centripetal force for the circular part of the motion of the man relative to the inertial frame. Likewise, the tangential component of the friction is responsible for the acceleration which the man has in the $-\hat e_2$ direction relative to the inertial frame.

I expect the centripetal component of force acting on the man (wrt the inertial frame) to have the form:

$$\vec X_c = -A \ \mathbf {\hat e_1}$$

Yes

And I expect the centripetal component of force acting on the man (wrt the non-inertial frame) to have the form:

$$\vec X_c = -A \ \mathbf {\hat e'_1}$$

In the non-inertial, rotating frame, the net force on the man is zero since the man has no acceleration in this frame. So, the net force in the $\hat e'_1$ direction is zero. The centripetal force due to friction is canceled by the fictitious centrifugal force.

Likewise, in the rotating frame, the component of the friction force in the $-\hat e'_2$ direction is canceled by the fictitious Coriolis force which is in the $+\hat e'_2$ direction.

The gravitational force is canceled by the normal force.

 

[JD_PM]

The thing to note is that as the man moves in his AM is decreasing $L = m\Omega r^2$.

Shouldn't be $L = m\Omega \Delta r^2$? (Where $\Delta r$ is the man's displacement)

Without any lateral forces, he should be spinning faster than the merry-go-round.

I am afraid I do not see what you mean. 'Without any lateral force' I guess you mean to neglect the friction force in the tangential direction.

 

[JD_PM]

By "true" I mean "actual" as opposed to "fictitious" (or "apparent").

I see it now.

The terms "Coriolis" and "centrifugal" do not apply to $\vec X$ because $\vec X$ is a true force while the Coriolis force and the centrifugal force are fictitious forces that are present only in the rotating frame. The true force $\vec X$ has a normal component: $+mg \hat e_3$, a centripetal component: $-m\Omega^2 x_1 \hat e_1$ , and a tangential component: $-2m \Omega V \hat e_2$. The vertical component is the normal force while the other two components represent the friction force acting on the man. The $\hat e_1$ component has a negative sign because this component of friction provides the necessary centripetal force for the circular part of the motion of the man relative to the inertial frame. Likewise, the tangential component of the friction is responsible for the acceleration which the man has in the $-\hat e_2$ direction relative to the inertial frame.

Now it is clear, thanks. By the way, didn't you forget the primes? :)

Mmm so with this exercise you tried to show me that the -ive sign in front of the $\mathbf {\hat e'_1}$ is due to the centripetal acceleration dragging the man inwards and that the -ive sign in front of the $\mathbf {\hat e'_2}$ is due to the tangential acceleration dragging the man to his left, didn't you?

Was this the aim of your exercise though?

 

[TSny]

Now it is clear, thanks. By the way, didn't you forget the primes? :)

Yes. But I was imagining that at the instant of time of interest, the axes of the two frames are coincident. So, $\hat e_1 = \hat e'_1$, etc.

Was this the aim of your exercise though?

I might have misunderstood your original question in post #1. I thought you were asking why the force $\hat X$ has a radial component in the negative $\hat e'_1$ direction instead of the positive $\hat e'_1$ direction. Similarly for the tangential component.

 

[PeroK]

Shouldn't be $L = m\Omega \Delta r^2$? (Where $\Delta r$ is the man's displacement)
I am afraid I do not see what you mean. 'Without any lateral force' I guess you mean to neglect the friction force in the tangential direction.

That's the correct expression for angular momentum about the centre. A purely radial force would reduce the distance $r$ while keeping $L$ constant. In which case the man's angular velocity must increase beyond $\Omega$.

That's why he needs a lateral force to maintain a constant angular velocity.