Rotational motion: playground spinning disk problem

[ChrisBrandsborg]

Homework Statement

A child is at a playground, and chooses to try the spinning disc (see figure). The radius of the disc is 2.00 m, and the coefficient of static friction between child-surface and disc-surface is μs = 0.350.
In the following questions, you must provide algebraic equations as well as final numbers.

a) If the child sits 1.50 m from the centre, how fast can the spinning disc turn, without her slipping off?

b) The evil big brother now spins the disc with ω = 2.00 rad/s. At what distance from the centre of the disc should the kid sit, to avoid falling off?

Having safely moved to the radius 0.500 m, the child starts complaining, becau- se she cannot get off the ride. The evil big brother decides to stop the rotating disc, by providing a constant tangential force.

c) How quickly can he stop the disc, without her sliding in any direction (neither forwards, backwards, inwards or outwards)?

Homework Equations

(possibly)[/B]
1. m⋅ν²/r (centrifugal force)

2. ω = ν/r (angular velocity)

3. a = -ν²/r (accelration)

3. F = ma (Newton Law)

The Attempt at a Solution

Relevant information:

r = 2.00m (radius of the disc)
μs = 0.350 (static friction)
Fs = μs⋅mg

r_a = 1.50m (distance from centre - in a)

My attempt:

mν²/r = μsmg

μs = ν_max²/rg

v_max = √μsrg

v_max = √0.350⋅1.50m⋅9.80m/s² = 2.27 rad/s

 

[gneill]

Hi ChrisBrandsborg,

Your homework help request might get more and better response if the title reflected the nature of the problem and the are of physics involved. I've changed the title to: Rotational motion: playground spinning disk problem.

You should be able to list a number of relevant equations including those related to rotational motion, centripetal acceleration, and friction. If you can't list them then perhaps you need to consult your text or course notes to review the topics. We really can't offer any help until you've shown that you've made an effort to research and attempt the problem.

 

[ChrisBrandsborg]

Hi ChrisBrandsborg,

Your homework help request might get more and better response if the title reflected the nature of the problem and the are of physics involved. I've changed the title to: Rotational motion: playground spinning disk problem.

You should be able to list a number of relevant equations including those related to rotational motion, centripetal acceleration, and friction. If you can't list them then perhaps you need to consult your text or course notes to review the topics. We really can't offer any help until you've shown that you've made an effort to research and attempt the problem.

Hi! I did write some relevant equations, as well as an attempt to solve it, but I am not sure about everything! Could really use some help!

 

[gneill]

In (a) the child sits at a radius of 1.50 m, so the centripetal force required will correspond to that radius. The speed of the disk should be specified in terms of rotational speed: radians per second (ω).

You may find it easier to work in angular quantities rather than translating back and forth to linear ones. Centripetal force, for example, is given by mω²r, bypassing the need to convert to linear speed to use mv²/r.

 

[ChrisBrandsborg]

In (a) the child sits at a radius of 1.50 m, so the centripetal force required will correspond to that radius. The speed of the disk should be specified in terms of rotational speed: radians per second (ω).

You may find it easier to work in angular quantities rather than translating back and forth to linear ones. Centripetal force, for example, is given by mω²r, bypassing the need to convert to linear speed to use mv²/r.

Yes, I forgot that it was rad/s. So the correct answer for a is: 2.27 rad/s ? :)

 

[gneill]

Yes, I forgot that it was rad/s. So the correct answer for a is: 2.27 rad/s ? :)

Nope. You can't just change the units on a velocity to make it angular velocity. How do you convert a linear velocity to an angular velocity?

Please don't keep changing your initial post. It makes the thread confusing to others when the followups don't reflect what came before.

 

[ChrisBrandsborg]

Nope. You can't just change the units on a velocity to make it angular velocity. How do you convert a linear velocity to an angular velocity?

Please don't keep changing your initial post. It makes the thread confusing to others when the followups don't reflect what came before.

Oh, okay, so I use mω²r = μsmg, and solve for ω?

 

[gneill]

Oh, okay, so I use mω²r = μsmg, and solve for ω?

That should do it

 

[ChrisBrandsborg]

That should do it

Okay, so:

ω_max = √(μsg/r) = 1.51 rad/s

Correct?

 

[gneill]

Yes. Looks good.

 

[ChrisBrandsborg]

I tried b as well:

Now ω = 2.00, and we want to find r.

μsmg = mω²r

r = μsg/ω² = 0.858m

Correct? :) Thanks a lot for taking your time!

 

[haruspex]

I tried b as well:

Now ω = 2.00, and we want to find r.

μsmg = mω²r

r = μsg/ω² = 0.858m

Correct? :) Thanks a lot for taking your time!

Yes.

 

[ChrisBrandsborg]

Then c:

r = 0.500m
Constant tangential force F_t
ω_stop = 0 rad/s
t = unknown

How do I find out how quickly he can stop the disc, without her sliding off?

 

[haruspex]

Constant tangential force Ft

Why would it be constant?

 

[ChrisBrandsborg]

Why would it be constant?

"The evil big brother decides to stop the rotating disc, by providing a constant tangential force"

 

[haruspex]

"The evil big brother decides to stop the rotating disc, by providing a constant tangential force"

Ok, I missed that.
What would be the angular deceleration?

 

[ChrisBrandsborg]

Ok, I missed that.
What would be the angular deceleration?

a = (ω_f - ω₀)/t ?

 

[haruspex]

a = (ω_f - ω₀)/t ?

No, I mean given an applied tangential force F_t.

 

[ChrisBrandsborg]

No, I mean given an applied tangential force F_t.

I am not sure what F_t is.. Is it: F_t = mg⋅sinθ?
So that a_t = gsin θ

if we find a, can we then use:

a = Δω/t, to find the time?

or is it a_t = rΔω/t

 

[haruspex]

I am not sure what Ft is

It is the applied tangential force you are trying to find. I used F_t because that is what you called it in post #13.
In terms of that unknown, what will the angular acceleration be?

 

[ChrisBrandsborg]

It is the applied tangential force you are trying to find. I used F_t because that is what you called it in post #13.
In terms of that unknown, what will the angular acceleration be?

I am not sure.. Can you help me? :)

 

[gneill]

The question states that you're looking for the minimum time to stop the rotation. Just assume that there's some angular acceleration α. What is the maximum value that α can have so that the girl doesn't slide? Once you have a value for α you can work on finding the time to stop.

 

[ChrisBrandsborg]

The question states that you're looking for the minimum time to stop the rotation. Just assume that there's some angular acceleration α. What is the maximum value that α can have so that the girl doesn't slide? Once you have a value for α you can work on finding the time to stop.

But which equation do I use to find a?

 

[gneill]

But which equation do I use to find a?

You're looking for the net force that the girl experiences. You've got centripetal acceleration and tangential acceleration operating, and friction force to oppose the resultant. Draw a diagram.

 

[ChrisBrandsborg]

You're looking for the net force that the girl experiences. You've got centripetal acceleration and tangential acceleration operating, and friction force to oppose the resultant. Draw a diagram.

a_c is towards the centre, and a_t and the friction is in the opposite direction of each other (and both perpendicular to a_c)?
Do I take the sum of those forces?

so ΣF = a_c + a_t -F_s ?

 

[gneill]

The friction force counters the tendency of the girl to continue moving in a straight line path (tangentially). This tendency is due to inertia.

To accomplish this, the friction force must supply a component to provide the required centripetal acceleration (radial) as well as one to counter the tangential acceleration due to the disk being slowed. Being components at right angles we naturally have a vector sum for the total friction required.

So you need to first write expressions for the tangential force that the girl experiences and the centripetal force that she experiences. Then sum them appropriately so that the sum can be compared to the maximal static friction force.

 

[ChrisBrandsborg]

The friction force counters the tendency of the girl to continue moving in a straight line path (tangentially). This tendency is due to inertia.

To accomplish this, the friction force must supply a component to provide the required centripetal acceleration (radial) as well as one to counter the tangential acceleration due to the disk being slowed. Being components at right angles we naturally have a vector sum for the total friction required.

So you need to first write expressions for the tangential force that the girl experiences and the centripetal force that she experiences. Then sum them appropriately so that the sum can be compared to the maximal static friction force.

But what are the expressions for the tangential force? The centripetal is mω²r, and the static friction force is: μsmg.

 

[gneill]

But what are the expressions for the tangential force? The centripetal is mω²r, and the static friction force is: μsmg.

You have an assumed angular acceleration $\alpha$. How is the instantaneous linear acceleration related to angular acceleration? What mass is being accelerated? So what's the force?

 

[ChrisBrandsborg]

You have an assumed angular acceleration $\alpha$. How is the instantaneous linear acceleration related to angular acceleration? What mass is being accelerated? So what's the force?

Instantaneous linear acc. = angular acc * radius ?
The spinning wheel is being accelerated? With the mass of gravity and the child?

F_t = mg⋅r⋅a ?

 

[gneill]

Instantaneous linear acc. = angular acc * radius ?
The spinning wheel is being accelerated? With the mass of gravity and the child?

F_t = mg⋅r⋅a ?

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.

 

[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?