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.