A person walks on a rotating platform at constant speed

[bunchedupwalrus]

Homework Statement

A large flat horizontal platform rotates at a constant angular speed ω. A person on the platform walks in a circular path of radius R₀ centered on the axis of the platform with a constant linear speed v relative to the platform’s surface. The coefficient of friction between the person and the platform’s surface is µ and the mass of the person is m. How fast can the person walk if: (a) they move in the direction of rotation? (b) if they move opposite the direction of rotation?

Homework Equations

Equation of motion for rotating body. ma'=Fext+Fcor+Fcentif
Fcor = -2mωXv'
Fcentrif = -m(ωX(ωXr'))
Fext= µmg
?a' = v²/r

ω = ω(khat) ( I chose counter clockwise rotation)
v = v(jhat)
r = R₀(ihat)

The Attempt at a Solution

Part a) I think my problem is with the a', I'm not sure if that is a valid substitution. Honestly pretty confused about how to set up the equation.

Subbing in the values, all of the m's cancel out and the unit vectors are all (ihat). I'm not entirely sure if the sign on my friction o a' is correct either:

a' = µg + 2ωv+ω²R₀
a' =v²/r
Subbing in the a, I get a quadratic for v, which factors to:

R₀µg=(v+ωR₀)², and that let's me solve for v.But is it right? This feels very shakey to me, the a' sub, the signs of the forces. I keep trying to wrap my head around it but it doesn't make proper sense.

I also treating this as a one dimensional problem, and get something entirely different. What is the best way to approach this problem?

Thanks in advance

 

[TSny]

Are you required to solve this in the rotating frame? If not, you might try analyzing it in the lab frame.

For your rotating frame analysis, I don't think you handled the signs properly. Suppose you take radially outward as the positive direction, so radially inward is negative.
What is the direction of the acceleration a'? Therefore, what is its sign?
What is the direction of the friction force? What is its sign?
Check the Coriolis and centrifugal forces to make sure you have the right signs for these.

Can you elaborate on treating the problem as one dimensional?

 

[bunchedupwalrus]

So if outwards is positive.

Centrifugal would be positive, friction negative, Coriolis I do not know, positive?

I don't know what the lefthand a is supposed to represent to be honest, and cannot guess it's sign. We only just went over this is in class, and now the semester is over. Textbook gives very little explanation.

Is the a substitution correct at least? I can try redoing it and paying better attention to the signs

 

[bunchedupwalrus]

The only tip given by the professor is to use polar coodinates.

 

[TSny]

So if outwards is positive.

Centrifugal would be positive, friction negative,

Yes.

Coriolis I do not know, positive?

The direction of the Coriolis force is determined by the direction of the cross product $-\vec{\omega} \times \vec{v}'$. So, make sure you can visualize the directions of $\vec{\omega}$ and $\vec{v}'$.

I don't know what the lefthand a is supposed to represent to be honest, and cannot guess it's sign.

The basic equation in the rotating frame is the vector equation: $m \vec{a}' = \vec{F}_{\rm ext} +\vec{F}_{\rm cor}+\vec{F}_{\rm cen}$.
$\vec{a}'$ is the acceleration vector of the object as seen in the rotating frame of reference. The magnitude and direction of $\vec{a}'$ for this problem can be determined by identifying the type of motion of the object relative to the rotating frame.

 

[bunchedupwalrus]

So a is zero, as he is moving at a constant velocity? Or do I use the v²/r as he is changing direction
I really appreciate the help, but still feel lost.

My new equation is
a'= -µg + 2(ωe_zXve_theta + ωe_zX(ωe_zXre_r)

Using polar coordinates
Which solves (if I am computing the crosses correctly) to

a'= -µg + 2ωv + ω²r all in the e_r direction

Rotating counterclockwise, person moving in counter clockwise direction

If a' is 0 I can solve for v, if it is v²/r I get a quadratic and can solve via factoring.

 

[TSny]

So a is zero, as he is moving at a constant velocity? Or do I use the v²/r as he is changing direction

Definitely the latter. The person is walking along a circle on the platform. So, if you were sitting on the platform, you would see the person moving with uniform circular motion. The acceleration $\vec{a}'$ is the acceleration associated with uniform circular motion. Thus, you should be able to state the magnitude and direction of the acceleration.

My new equation is
a'= -µg + 2(ωe_zXve_theta + ωe_zX(ωe_zXre_r)

OK, but the directions for the term on the left and the first term on the right are not clearly indicated here.

Using polar coordinates
Which solves (if I am computing the crosses correctly) to

a'= µg + 2ωv + ω²r all in the e_r direction

What can you substutute for a' (including the sign)?
Did you get the sign of µg right?

 

[bunchedupwalrus]

Definitely the latter. The person is walking along a circle on the platform. So, if you were sitting on the platform, you would see the person moving with uniform circular motion. The acceleration $\vec{a}'$ is the acceleration associated with uniform circular motion. Thus, you should be able to state the magnitude and direction of the acceleration.

It should be negative as it is pulling the person in towards the center?OK, but the directions for the term on the left and the first term on the right are not clearly indicated here.
I don't understand sorry. Omega is pointing out of the page, as it is rotating CC, so e_z. eTheta is moving CC, inline with V

What can you substutute for a' (including the sign)?
Did you get the sign of µg right?
I missed it on the second one, fixed now.
-v²/r ?

 

[TSny]

It should be negative as it is pulling the person in towards the center?

Yes.

TSny: OK, but the directions for the term on the left and the first term on the right are not clearly indicated here.

bunchedupwalrus:I don't understand sorry. Omega is pointing out of the page, as it is rotating CC, so e_z. eTheta is moving CC, inline with V

In the equation a'= -µg + 2(ωezXvetheta + ωezX(ωezXrer), the term on the left side is written as a' without a unit vector to show direction. Or maybe you meant a' to be the acceleration vector? It's hard to tell. Symbols for vectors are usually written with an arrow over them or else they are written in bold face type. Likewise, the first term on the right side is written as -µg without a unit vector to show direction. Maybe you meant -µg e_r. I guess I'm being picky, but clarity is important.

-v²/r ?

OK. The minus sign indicates radially inward. So, when you solve your equation for μ, do all of the signs work out as they should?

 

[bunchedupwalrus]

Yes.

In the equation a'= -µg + 2(ωezXvetheta + ωezX(ωezXrer), the term on the left side is written as a' without a unit vector to show direction. Or maybe you meant a' to be the acceleration vector? It's hard to tell. Symbols for vectors are usually written with an arrow over them or else they are written in bold face type. Likewise, the first term on the right side is written as -µg without a unit vector to show direction. Maybe you meant -µg e_r. I guess I'm being picky, but clarity is important.

OK. The minus sign indicates radially inward. So, when you solve your equation for μ, do all of the signs work out as they should?

I understand about the unit vectors, I just couldn't figure out how to express them in the forum, sorry about that.

a' was meant to be acceleration vector from ma'

If I go with

-v²/R₀ = -µg + 2ωv+ω^2R₀ (*All in e_r direction)

I end up with R₀µg = (v+ωR₀)²

Which gives a v_max of sqrt(R₀µg)-2ωR₀

 

[bunchedupwalrus]

My bad sorry, it becomes (v+ωR₀)^2, have corrected it.

 

[TSny]

-v²/R₀ = -µg + 2ωv+ω^2R₀ (*All in e_r direction)

Looks good.

I end up with R₀µg = (v+ωR₀)²

Which gives a v_max of sqrt(R₀µg)-2ωR₀

OK. Check the last term. Not sure the factor of 2 should be there.