Rotational motion: playground spinning disk problem

[ChrisBrandsborg]

Don't look at gravity yet. That will come in when you look at the friction force itself. For now deal with the radial and tangential forces which are not vertical and do not involve gravity.

Only the mass of the child is of interest here. We don't know how massive the disk is or what its moment of inertia might be. That's the Bully's problem to deal with as he musters enough force to cause some acceleration $\alpha$.

The disk is accelerating (slowing down) and for the child to remain in the same location on it she must slow down too. So friction provides the force between the disk and her to accomplish this. By Newton's 3rd law the force accelerating her is matched by the so-called inertial force of her mass resisting acceleration.

View attachment 106580

Okay, so F_t = m_child⋅ra, and then we sum all the forces together to find a?

 

[gneill]

Okay, so F_t = m_child⋅ra, and then we sum all the forces together to find a?

You definitely want to sum the components to find the total force required to keep the child pinned in place on the disk. How will you sum these components?

By the way, you can insert various Greek characters and other math symbols from a menu that will appear if you select the $\Sigma$ icon on the edit window icon bar. It's best to keep angular "α" distinct from linear "a"

 

[ChrisBrandsborg]

You definitely want to sum the components to find the total force required to keep the child pinned in place on the disk. How will you sum these components?

By the way, you can insert various Greek characters and other math symbols from a menu that will appear if you select the $\Sigma$ icon on the edit window icon bar. It's best to keep angular "α" distinct from linear "a"

∑F = F_t + F_c - F_s
ΣF = mrα + mω²r - μsmg

 

[ChrisBrandsborg]

I will take a break for today! I will continue tomorrow! Would love some more guidence though :D

 

[gneill]

∑F = F_t + F_c - F_s
ΣF = mrα + mω²r - μsmg

Remember, you're dealing with vector components here. How do you sum vector components to find the magnitude of the vector?

I will take a break for today! I will continue tomorrow! Would love some more guidence though :D

Sure, no worries.

 

[haruspex]

∑F = Ft + Fc - Fs

No.
Centripetal force is not an applied force. It is that resultant of the applied forces which leads to the centriptal acceleration:
$\vec F_c=\Sigma\vec F=\vec F_t+\vec F_s$

I note that there has been a change of notation in the thread. Originally you defined F_t as the force applied by the bully. Now it is being used for the tangential force applied to the girl, and the bully is applying force F_bully. That's ok, just as long as we all understand that.

ΣF = mrα + mω2r - μsmg

The static frictional force only equals μ_s multiplied by the normal force (mg here) when just about to slip. Of course, here you are interested in the case where it is about to slip so the equation is valid; just pointing out that it is not in general true.

Also, you may be confused that gneill seems to be taking quite a different approach from mine. It's just a matter of the order of steps. At some point, we have to connect F_bully with F_t, and connect F_t with α. I was starting with the first of those, whereas gneill has started on the second. Both have to be done.

 

[ChrisBrandsborg]

No.
Centripetal force is not an applied force. It is that resultant of the applied forces which leads to the centriptal acceleration:
$\vec F_c=\Sigma\vec F=\vec F_t+\vec F_s$

I note that there has been a change of notation in the thread. Originally you defined F_t as the force applied by the bully. Now it is being used for the tangential force applied to the girl, and the bully is applying force F_bully. That's ok, just as long as we all understand that.

The static frictional force only equals μ_s multiplied by the normal force (mg here) when just about to slip. Of course, here you are interested in the case where it is about to slip so the equation is valid; just pointing out that it is not in general true.

Also, you may be confused that gneill seems to be taking quite a different approach from mine. It's just a matter of the order of steps. At some point, we have to connect F_bully with F_t, and connect F_t with α. I was starting with the first of those, whereas gneill has started on the second. Both have to be done.

Yeah, I forgot that F_c is not really a force.
So what should I do now? What is the next step?

 

[haruspex]

Yeah, I forgot that F_c is not really a force.
So what should I do now? What is the next step?

As I posted, you have a choice of which part to do next.
You need to relate the applied force, F_bully, to the angular acceleration of the system. That is probably the easier part. What would that relationship be?
You also need to relate the angular acceleration of the system to the tangential acceleration of the girl, and thus to the total acceleration of the girl.

 

[ChrisBrandsborg]

As I posted, you have a choice of which part to do next.
You need to relate the applied force, F_bully, to the angular acceleration of the system. That is probably the easier part. What would that relationship be?
You also need to relate the angular acceleration of the system to the tangential acceleration of the girl, and thus to the total acceleration of the girl.

So first, we need to relate F_bully with the angular acceleration, and then use that to find the total acc. of the girl?
F_bully is the opposite direction of the motion, but we don´t know what it is? How do we find it?
And the angular acceleration, is that = rα?

or wait... F_bully = -mrα, but what is then the angular acceleration?

 

[haruspex]

Fbully is the opposite direction of the motion, but we don´t know what it is? How do we find it?

We are trying to find equations that relate variables. We do not care yet about finding values for any of them. F_bully is the thing we are trying to find, so the value for that will come right at the end.

Fbully = -mrα

Since we are not told the mass of the disc, that's almost right. But the bully is not pushing directly on the girl, the bully is pushing at the edge of the disc. Does the word "torque" ring any bells?

 

[ChrisBrandsborg]

We are trying to find equations that relate variables. We do not care yet about finding values for any of them. F_bully is the thing we are trying to find, so the value for that will come right at the end.

Since we are not told the mass of the disc, that's almost right. But the bully is not pushing directly on the girl, the bully is pushing at the edge of the disc. Does the word "torque" ring any bells?

Yes, so it should be: F_bully = mr²α (where m is the mass of the point he pushes?)

 

[Kaura]

Remember that the force keeping the child from slipping off from the ride is friction
If friction and centripetal force are equal then what will happen to the child?

 

[ChrisBrandsborg]

Remember that the force keeping the child from slipping off from the ride is friction
If friction and centripetal force are equal then what will happen to the child?

Then she will keep spinning around with constant angular speed?

 

[Kaura]

Then she will keep spinning around with constant angular speed?

Yes when the force of friction is greater than or equal to the centripetal force then the angular velocity of the child will not change as the centripetal force is not strong enough to accelerate the child

 

[ChrisBrandsborg]

Yes when the force of friction is greater than or equal to the centripetal force then the angular velocity of the child will not change as the centripetal force is not strong enough to accelerate the child

Yes, so where do we go from here?

 

[ChrisBrandsborg]

I need to finish this before tomorrow. I still need help to solve this :)

 

[jbriggs444]

Yes, so it should be: F_bully = mr²α (where m is the mass of the point he pushes?)

This is a dead end. The force with which the bully pushes does not figure into the problem or its solution. You do not need to know it.

What matters is the angular acceleration that the bully achieves. That is the parameter that you care about, not how it is attained.

 

[ChrisBrandsborg]

This is a dead end. The force with which the bully pushes does not figure into the problem or its solution. You do not need to know it.

What matters is the angular acceleration that the bully achieves. That is the parameter that you care about, not how it is attained.

Can you just write how you would have started to solve this, from the beginning.. I am really confused now!

 

[jbriggs444]

Can you just write how you would have started to solve this, from the beginning.. I am really confused now!

That's not how we roll here. It is your problem to solve. We give hints, not complete solutions.

Edit: That said, you've had a lot of hints that point you in the right direction.

You need to find the angular acceleration that will allow the child to just barely remain in place while slowing down.

Then you need to find the amount of time it will take for that angular acceleration to stop the merry-go-round.

 

[ChrisBrandsborg]

That's not how we roll here. It is your problem to solve. We give hints, not complete solutions.

Not complete.. Just start over, and help me on the right track. What do we know? And what do I need to do first?

 

[jbriggs444]

Not complete.. Just start over, and help me on the right track. What do we know? And what do I need to do first?

@gneill has you on the right track. @haruspex is leading you on a wild goose chase for a parameter, F_bully, that is irrelevant. (*)

In posts #30 through #35, @gneill is trying to get you to remember how to add vectors.

(*) The F_bully is only relevant in that we are told that it is fixed. The moment of inertia of the carousel with an unmoving child is fixed. A fixed force applied at a fixed radius on an object with a fixed moment of inertia will result in a fixed torque and a fixed angular acceleration. Other than allowing us to infer a constant angular acceleration, we can ignore F_bully entirely.

 

[gneill]

Let's pause for a moment to recap.

The current situation is that the child is located at radius r = 0.500 m, the disk and child share rotational velocity ω = 2.00 rad/sec, and BB (Big Bully) is causing a fixed angular acceleration (currently unknown magnitude) α in order to stop the motion as quickly as possible.

The most expedient way forward is to draw a diagram showing the forces acting. Note that these will be Cartesian vectors, or at least vector components. So you'll need to be able to convert angular quantities to linear ones as required. Recognize that the centripetal force is provided by friction, as is the tangential force that keeps the girl's linear (and angular) speed equal to that of the disk at her location on it.

Then write expressions for the magnitudes of these forces. Then write an expression combining those into one equation that needs to be satisfied.

So let's start. For the child:

What is an expression for the magnitude of the centripetal force that must be operating? $f_c = ~~?$

What is an expression for the magnitude of the tangential force? $f_t = ~~?$

What is an expression for the magnitude of the maximal static friction force? $f_f = ~~?$

Now, how are they related in this problem? Remember that these are vector magnitudes. You should be able to pencil them in next to the force vectors that you drew on your diagram.

 

[haruspex]

. @haruspex is leading you on a wild goose chase for a parameter, Fbully, that is irrelevant. (*)

Ouch! Thanks for pointing that out.
Very sorry, @ChrisBrandsborg. I had picked up on your reference to the force exerted by the bully at some post near the start and not gone back to check what the question actually asked for.

 

[ChrisBrandsborg]

Ouch! Thanks for pointing that out.
Very sorry, @ChrisBrandsborg. I had picked up on your reference to the force exerted by the bully at some post near the start and not gone back to check what the question actually asked for.

hehe, no worries :)

 

[ChrisBrandsborg]

Let's pause for a moment to recap.

The current situation is that the child is located at radius r = 0.500 m, the disk and child share rotational velocity ω = 2.00 rad/sec, and BB (Big Bully) is causing a fixed angular acceleration (currently unknown magnitude) α in order to stop the motion as quickly as possible.

The most expedient way forward is to draw a diagram showing the forces acting. Note that these will be Cartesian vectors, or at least vector components. So you'll need to be able to convert angular quantities to linear ones as required. Recognize that the centripetal force is provided by friction, as is the tangential force that keeps the girl's linear (and angular) speed equal to that of the disk at her location on it.

Then write expressions for the magnitudes of these forces. Then write an expression combining those into one equation that needs to be satisfied.

So let's start. For the child:

What is an expression for the magnitude of the centripetal force that must be operating? $f_c = ~~?$

What is an expression for the magnitude of the tangential force? $f_t = ~~?$

What is an expression for the magnitude of the maximal static friction force? $f_f = ~~?$

Now, how are they related in this problem? Remember that these are vector magnitudes. You should be able to pencil them in next to the force vectors that you drew on your diagram.

Thanks for taking your time! Easier to think when I have the useful information set up like this. I will look at it later today :)