Max distance to place an object so it won't slide (torque and friction)

[Femme_physics]

Homework Statement

http://img196.imageshack.us/img196/3732/31591397.jpg

An area slanted at an angle of 30 degrees turns aroun a vertical axis as depicted in drawing. The period time of the rotational motion is 3 seconds. On the surface is an object whose mass is 6kg.
The static friction coeffecient between the body and the surface is 0.4

Calculate at which max distance (from the rotation axis) can the object be placed so it won't slide up the surface?

The Attempt at a Solution

I wasn't really sure what to do! How does the torque on the surface affects the forces in my static equations? I just drew a free body diagram and tried solving for stuff randomly. I've no idea how to approach the solution to this. Any help/hints?

http://img405.imageshack.us/img405/8708/80167579.jpg

 

[I like Serena]

I wasn't really sure what to do! How does the torque on the surface affects the forces in my static equations? I just drew a free body diagram and tried solving for stuff randomly. I've no idea how to approach the solution to this. Any help/hints?

You're on the right track with your FBD!

But you didn't account yet for the centripetal acceleration.
The 3 forces you drew have to add up to ma.

You also didn't mark the distance from the rotation axis yet, which is what you have to solve.


Btw, which torque on the surface do you mean?

 

[ehild]

You want the mass move along a horizontal circle with given angular velocity. What has to be the resultant of all forces?
Remember N is a force of constraint. It is equal to the normal component of gravity if the slope is in rest, but it turns now...

ehild

Ohhh, ILS beat me

 

[Femme_physics]

Btw, which torque on the surface do you mean?

I mean centripetal acceleration (apparently!), sorry :)

You're on the right track with your FBD!

w00t! (oh no, smiley infection is moving topics!)

You also didn't mark the distance from the rotation axis yet, which is what you have to solve.

This is another weird thing. They didn't write any distances or measurements! So that's another bump in the road that I'm not sure how to maneuver through.

(illustration to help analogy)

http://img810.imageshack.us/img810/529/bumproad.jpg --------------------------

Okay, first things first, I need to find centripetal acceleration. V squared/R. Tricksy, since I don't have R. This entire thing is so confusing without ANY given measurements at all. As far as I'm concerned, this thing can be in nanometers. Okay, yea, so the object is 6kg so it doesn't make sense. Still, shouldn't they have written distances?

You want the mass move along a horizontal circle with given angular velocity. What has to be the resultant of all forces?
Remember N is a force of constraint. It is equal to the normal component of gravity if the slope is in rest, but it turns now...

I really lost you here, especially in the first sences: "You want the mass move along a horizontal circle with given angular velocity. What has to be the resultant of all forces? "

Can you elaborate for me a bit, please?

Ohhh, ILS beat me

He's fast isn't he?

 

[ehild]

I really lost you here, especially in the first sences: "You want the mass move along a horizontal circle with given angular velocity. What has to be the resultant of all forces? "

Can you elaborate for me a bit, please?

The wedge or what turns round, and the mass moves together with it. As it does not slide up or down, it moves at constant height, does not it? The period of rotation is given so you can get the angular speed.

He's fast isn't he?

No, Physicsforums was too slow, it just refused to accept my answer, it does such things sometimes...

 

[Femme_physics]

it moves at constant height

If it MOVES on an inclined plane, how can it be at a constant height?

No, Physicsforums was too slow, it just refused to accept my answer, it does such things sometimes...

I got 504 gateway timeout myself if this is what you're referring to. :-/

 

[I like Serena]

He's fast isn't he?

Well, this time I was slow!
But then, I took my time enjoying the drawing and the smileys and all!
I've just been sitting here, dreaming away!

This is another weird thing. They didn't write any distances or measurements! So that's another bump in the road that I'm not sure how to maneuver through.

Okay, first things first, I need to find centripetal acceleration. V squared/R. Tricksy, since I don't have R. This entire thing is so confusing without ANY given measurements at all. As far as I'm concerned, this thing can be in nanometers. Okay, yea, so the object is 6kg so it doesn't make sense. Still, shouldn't they have written distances?

So mark a horizontal distance from the block to the axis of rotation.
There! You have R, which is also the distance they ask you to solve.

Together with all the other measurements (mass, angle, coefficient of friction, frequency), you have all you need!

 

[Apophilius]

The point at which the object starts slipping is the point at which you no longer have any additional forces with which to apply a greater centripetal force. Remember, you don't know r because what you're trying to solve for is r.

Applying these equations should get you rolling:

w = 2pi / T where T is your period.
ac = rw^2 where ac is your centripetal acceleration, and w is angular velocity. You could also find v, and us ac = v^2 / r, whatever you prefer.

Something to keep in mind: you're rotating horizontally, not parallel to the plane the block's on, so keep your trig straight. Not that you'll have trouble with this, it's just some stuff I like to remember when solving problems.

See if you can't get going with that.

 

[ehild]

If it MOVES on an inclined plane, how can it be at a constant height?

The mass does not move along the plane. The question is what the maximum distance is from the axis so it does not move up. The body is in a rotating frame of reference and feels a centrifugal force pointing away from the axis. That has a component along the slope. I do not like accelerating frames of reference (neither do you, I assume), but without static friction the body would move with respect to the slope not only up or down, but sideways, too. So we assume that the body is in rest with respect to the slope and do "reverse engineering" to find the forces necessary to that.

I got 504 gateway timeout myself if this is what you're referring to. :-/

That is its custom, and I have lost my letters quite often. Now I know what to do, copy my answer and go out from PF and try again a little later.

ehild

 

[I like Serena]

That is its custom, and I have lost my letters quite often. Now I know what to do, copy my answer and go out from PF and try again a little later.

Actually, I found I it suffices to click "refresh", and when asked to resend, click "Resend".
Wait a couple of minutes and try again if you get another gateway timeout.

 

[ehild]

Actually, I found I it suffices to click "refresh", and when asked to resend, click "Resend".
Wait a couple of minutes and try again if you get another gateway timeout.

thanks, but I am usually get bored with PF if it is so idle and go somewhere else. I do not have much time to wait

ehild

 

[Femme_physics]

So mark a horizontal distance from the block to the axis of rotation.
There! You have R, which is also the distance they ask you to solve.

Together with all the other measurements (mass, angle, coefficient of friction), you have all you need!

Actually, if I'm looking at the formula for centripetal acceleration, I don't have V and I don't have R. So how can I solve for centripetal acceleration?

I've just been sitting here, dreaming away!

*covers you with a blanket*

That is its custom, and I have lost my letters quite often. Now I know what to do, copy my answer and go out from PF and try again a little later.

Oh yea, been doing that since 2001. I don't be trusting this "interweb" technewlogy...

thanks, but I am usually get bored with PF if it is so idle and go somewhere else. I do not have much time to wait

Are you saying I'm not posting enough questions?

Rest be assured, this is my folder, each file contains 7 questions :)

http://img863.imageshack.us/img863/7837/seethisf.jpg I'll be spending my June here till I MASTER this stuff (master it like splinter from Ninja turtles!)^^
Though, hopefully I'll figure most of it on my own and won't have to poke you guys. I actually had a 5 questions winning streak before flunking with this one. Sorry for going off-topic! :) My next reply will be about the exercise provided I have no more distractions.:shy:

So, Apop, ehild, ILS, thanks. Will get to it :)

 

[I like Serena]

Actually, if I'm looking at the formula for centripetal acceleration, I don't have V and I don't have R. So how can I solve for centripetal acceleration?

It's the third dimension!

The block is actually coming out of the page hurtling toward you!
And just before it touches you, it turns around, dives into and through the page, completing its circle!
And it that's not enough, it only takes 3 seconds to do all that!

Its speed is v = ω x R

Its centripetal acceleration is a = v² / R = ω² R

Sorry for going off-topic! :) My next reply will be about the exercise provided I have no more distractions.:shy:

Oh really!?

 

[Femme_physics]

The block is actually coming out of the page hurtling toward you!

*swings arms defensively!*

And just before it touches you, it turns around, dives into and through the page, completing its circle!

Phew. I really should take a little more distance from my problems, next time! Of course, if I had any measurements... *grumbles*...

Its speed is v = ω x R

1 equation, two unknowns

Oh really!?

Really

The point at which the object starts slipping is the point at which you no longer have any additional forces with which to apply a greater centripetal force. Remember, you don't know r because what you're trying to solve for is r.

Applying these equations should get you rolling:

w = 2pi / T where T is your period.

I did solve for omega

ac = rw^2 where ac is your centripetal acceleration

2 unknowns, 1 equation!

The mass does not move along the plane. The question is what the maximum distance is from the axis so it does not move up.

I might have mistranslated. It said "so that the object won't move up the plane"

 

[ehild]

It said "so that the object won't move up the plane"

Yes, I know, but it is weird if moving down is allowed. In this case it gets closer to the axis, but you want the maximum distance. And the coefficient of static friction is given, that suggests no motion with respect to the plane. Agree?

ehild

 

[I like Serena]

*swings arms defensively!*

Phew. I really should take a little more distance from my problems, next time! Of course, if I had any measurements... *grumbles*...

I guess if it's in nanometers, there's no risk of the block hitting you!

But then, if it's in meters...

1 equation, two unknowns

I did solve for omega

2 unknowns, 1 equation!

Hmm, let's see...

Sum of vertical forces = 0
Sum of horizontal forces = m a
a = ω² R
Fs = 0.4 x N

Unknowns are: N, Fs, a, and R.

That is...

4 equations with 4 unknowns!

And you can probably fill a couple of those in, one after the other!


Oh, I know, you just don't want to do the work!

 

[Femme_physics]

Yes, I know, but it is weird if moving down is allowed. In this case it gets closer to the axis, but you want the maximum distance. And the coefficient of static friction is given, that suggests no motion with respect to the plane. Agree?

I agree there's no motion at the Y axis! Yes, I do.

4 equations with 4 unknowns!

Oh, god, one of those...

Oh, I know, you just don't want to do the work!

True, I do shy away from multiple equations and multiple unknown problems. They're time consuming, and less about physics, more about math, and waste half my day trying to solve them. :P

I do have a calcular function for 3 eq 3 unknowns, but I don't think I have one for 4 eq 4 unknowns. *pouts*

Alright. I'll get cracking after diving my head in sand.^^

Thank you :)

 

[ehild]

Do not let you frighten away by any number of unknowns. After all, you are student of engineering are not you?
Just write up all horizontal and vertical force components, and find those two equations for the horizontal components and the vertical ones. Basically, you have two unknowns, N and R, as you can take Fs=0.4N, and parallel to the slope. The centripetal force contains only R as unknown, as it is Rw^2 and you know the angular speed. Go ahead, show your equations.

ehild

 

[Femme_physics]

I wouldn't really call it 4 equations 4 unknowns. Just a bit of untangling. 4 equations 4 unknowns is IMO when you can't get any numeral result (without an unknown) without using all the 4 equations. Just IMHO. :)

Is that it?

http://img821.imageshack.us/img821/4932/requals.jpg
((how I got to the other results is in the first attempt))

(but you're right, I shouldn't be afraid of multiple equations. I'm not, really!)

 

[I like Serena]

Oh, god, one of those...

I'm afraid I haven't been able to protect you from them!
But I'll try to help you confront them!

True, I do shy away from multiple equations and multiple unknown problems. They're time consuming, and less about physics, more about math, and waste half my day trying to solve them. :P

I do have a calcular function for 3 eq 3 unknowns, but I don't think I have one for 4 eq 4 unknowns. *pouts*

Alright. I'll get cracking after diving my head in sand.^^

Thank you :)

Well, with ehild's suggestion I think it would fit in your calculator!

 

[I like Serena]

I wouldn't really call it 4 equations 4 unknowns. Just a bit of untangling. 4 equations 4 unknowns is IMO when you can't get any numeral result (without an unknown) without using all the 4 equations. Just IMHO. :)

Is that it?

http://img821.imageshack.us/img821/4932/requals.jpg
((how I got to the other results is in the first attempt))

(but you're right, I shouldn't be afraid of multiple equations. I'm not, really!)

No, this is not quite it.

You're using an N that should be a little bigger, since it also has to contribute to the resulting centripetal acceleration.
Fs will be a little bigger then as well, also contributing to the centripetal acceleration.

I'm not quite clear what you did with Σ F_x here, but I suspect you did not account for the fact that the centripetal acceleration is horizontal and not along the slope, which you defined as the x axis.EDIT: Btw, the measurement unit is not quite unknown, since you have the units Newtons, kilograms, and seconds.
If you consider that a Newton is actually a derived unit in kilograms, meters, and seconds, you'll see that your measurement unit for R will be meters!

 

[Femme_physics]

You're using an N that should be a little bigger, since it also has to contribute to the resulting centripetal acceleration.
Fs will be a little bigger then as well, also contributing to the centripetal acceleration.

I'm confused, isn't N just

Sum of all forces on Y = -Wcos(30)+N = 0Therefor I should have the correct figure for N, no?

 

[I like Serena]

I'm confused, isn't N just

Sum of all forces on Y = -Wcos(30)+N = 0


Therefor I should have the correct figure for N, no?

No, you left out the needed centripetal acceleration.

Perhaps you should think of it this way.

The centrifugal pseudo force presses the block extra hard against the slope.
This generates an increase in normal force...

If you would increase the circular speed, the centrifugal force will make the block slide up the slope (until it falls over the edge! )

 

[Femme_physics]

In order to find centrifugal force, I need R, in order to find R, I need N, which is more than W cos(30) since I have to consider the centrifugal addition to the normal force. Great, I'm in a dead end, going in circles :(

 

[ehild]

In order to find centrifugal force, I need R, in order to find R, I need N, which is more than W cos(30) since I have to consider the centrifugal addition to the normal force. Great, I'm in a dead end, going in circles :(

I think you have studied systems of equations with two or more variables. Just find out the relation between the forces. You are a clever girl, you can do it.

ehild

 

[Femme_physics]

I want to prove your confidence in me right, I really do!

Sum of all forces on Y = -Wcos(30)+N = 0
Sum of all forces on X = Fs + Wcos(60) = 0
Fs = N x 0.4
a = v squared/R
a = omega squared times R


Is this everything I need to solve? ILS wrote to me

Sum of vertical forces = 0
Sum of horizontal forces = m a

But he didn't write what those forces are! What I wrote my attempts, he told me I'm wrong, now I'm just losing the string.

Well, I'll be going on the bus back home soon so I'll probably tackle this exercise in only a few hours from now. Thanks^^

REALLY appreciate all your help! :)

 

[ehild]

Think: The object accelerates. How can it be if the sum of all forces is zero? ehild

 

[Apophilius]

I'll say it one more time: your period, which is given to you, is related to omega (w) through w = 2pi / T where T is your period. So you can find w.

Centrifugal force is a false force. You won't include it in your free body diagrams at all. As for centripetal force, it is only caused by existing forces (friction, normal, etc), it is in itself not an actual independent force.

What you want to do is set your centripetal force (Fc = mrw^2) equal to all of the horizontal force components (this is where your free body diagram will help). You can find all of those force components. Thereafter you will have a system of equations with only one variable unaccounted for, that variable being r. R is the horizontal distance from the axis of rotation, and using the fact that you have a 30 degree angle, you can find the block's position on the incline using trig.

 

[I like Serena]

In order to find centrifugal force, I need R, in order to find R, I need N, which is more than W cos(30) since I have to consider the centrifugal addition to the normal force. Great, I'm in a dead end, going in circles :(

Yes, the block is going around in circles.
And if you sit on it, you'll probably get dizzy!

Is this everything I need to solve? ILS wrote to me

But he didn't write what those forces are! What I wrote my attempts, he told me I'm wrong, now I'm just losing the string.

Well, I'm trying to let you wrestle a bit on your own with this problem.
Takes a bit of effort though, because I'd like to give you the world, and not just this block.

Let me give you a hint anyways.
How does "a" figure in your sums of forces?

Well, I'll be going on the bus back home soon so I'll probably tackle this exercise in only a few hours from now. Thanks^^

REALLY appreciate all your help! :)

See you later then!

 

[Femme_physics]

I'll say it one more time: your period, which is given to you, is related to omega (w) through w = 2pi / T where T is your period. So you can find w.

That I found even before posting the problem-- as a proof you can see my original post where I show that I found omega!

Centrifugal force is a false force. You won't include it in your free body diagrams at all. As for centripetal force, it is only caused by existing forces (friction, normal, etc), it is in itself not an actual independent force.

OHHHHHHHHHHHHHHHHHHHHHHH! Now I get the meaning of why ILS keeps calling it a pseudo-force :)

What you want to do is set your centripetal force (Fc = mrw^2) equal to all of the horizontal force components (this is where your free body diagram will help).

Why am I doing it? centripetal force is ma, yes? So I'm doing sum of all forces on x = ma, which is exactly what I've done! is this just sum of all forces = ma?

Thereafter you will have a system of equations with only one variable unaccounted for, that variable being r. R is the horizontal distance from the axis of rotation, and using the fact that you have a 30 degree angle, you can find the block's position on the incline using trig.

Well, I know the equation to get R but I was told that my normal force was wrong in that equation. But I do know the final step once I have the right normal force.

Yes, the block is going around in circles.
And if you sit on it, you'll probably get dizzy!

I don't need anymore help getting dizzy right about now!

Well, I'm trying to let you wrestle a bit on your own with this problem.

Com'here *grabs the triangle and does a trapping headbutt followed by a battering ram, dropkick and an eye poke*

Rawr!1

How does "a" figure in your sums of forces?

Can you rephrase that, please? I don't understand

 

[I like Serena]

Centrifugal force is a false force. You won't include it in your free body diagrams at all. As for centripetal force, it is only caused by existing forces (friction, normal, etc), it is in itself not an actual independent force.

OHHHHHHHHHHHHHHHHHHHHHHH! Now I get the meaning of why ILS keeps calling it a pseudo-force :) .

Actually, there's a little more to it than that.
That's when we would talk about rotating non-inertial frames, in which the centrifugal force is an actual independent force (just like gravity ;)).
This problem could be solved like that as well.
But I prefer to not make Fp dizzier than she already is!

Com'here *grabs the triangle and does a trapping headbutt followed by a battering ram, dropkick and an eye poke*

Rawr!1

Not so fast!
*make head-roll over the ground avoiding the trapping headbutt and the battering ram, but is floored by the dropkick*
*sticks up fork-hand to ward off eye poke, eye poke is averted, but it hurts!*
*rolls backward and jumps limberly onto feet*
*makes the Matrix hand gesture, saying com'on grrl... (hiding the hurt of the averted eye poke)*

Why am I doing it? centripetal force is ma, yes? So I'm doing sum of all forces on x = ma, which is exactly what I've done!

is this just sum of all forces = ma?

Can you rephrase that, please? I don't understand

You're saying that you've done the sum of all forces on x = ma, but you haven't.
Because in your force sum formulas, there's no "ma".

 

[kjohnson]

OHHHHHHHHHHHHHHHHHHHHHHH! Now I get the meaning of why ILS keeps calling it a pseudo-force :)

Think of the centripal force as the inertial part of the equation...the 'ma' side. So what you are doing is summing all the forces that point in the radial direction (pointing into the circle) and setting that equal to the centrepital force.

Why am I doing it? centripetal force is ma, yes? So I'm doing sum of all forces on x = ma, which is exactly what I've done!

is this just sum of all forces = ma?

Well in a way yes the sum of the forces is = ma...
BUT instead of 'a' being just a constant value (like gravity) or a random function (a=5t) it is equal to rw^2. This rw^2 term is completely determined by the fact that the path of motion is a circle. And the direction of this rw^2 is always pointing into the center of the circle as the body rotates.

The centepital force then makes up the entire inertial side and is =ma=mrw^2

 

[ehild]

What about getting started at last?
It is better to use horizontal (x) and vertical (y) components of forces than normal and tangential ones, as the resultant force is horizontal, and the only force we know -gravity- is vertical. You have to choose the positive direction both for x and y.
So:
N is the magnitude of the normal force, it is just N, no matter what mg is. What are its horizontal and vertical components in terms of the angle of the slope?
Fs is the static friction and its direction is so that it prevents the body from upward motion, according to the problem text, so the force of friction points downward.
What are its vertical and horizontal components?

When you have all these components of forces, collect the horizontal ones which result in the centripetal force, and collect all the vertical ones, which make zero.

Could you please show what you get?

ehild

 

[Femme_physics]

You're saying that you've done the sum of all forces on x = ma, but you haven't.
Because in your force sum formulas, there's no "ma".

Typo. :-/ Should be ma not 0.

Think of the centripal force as the inertial part of the equation...the 'ma' side. So what you are doing is summing all the forces that point in the radial direction (pointing into the circle) and setting that equal to the centrepital force.

Centrifugal force is a false force. You won't include it in your free body diagrams at all. As for centripetal force, it is only caused by existing forces (friction, normal, etc), it is in itself not an actual independent force.

Then how come I'm not considering moment of inertia, friction, etc, or whatever in my calculations of centripetal force? Shouldn't it matter if those are the things that REALLY cause centripetal force?

In kinematics/dynamics, we ignore friction by saying "smooth surface" and ignore moment of inertia by saying "point mass". I'm curious why aren't we told anything in questions about moment of inertia and friction if it so heavily affects the score. What if there was no friction, indeed? Would there be no centripetal force/much much lesser to an insignificant level?

Not so fast!
*make head-roll over the ground avoiding the trapping headbutt and the battering ram, but is floored by the dropkick*
*sticks up fork-hand to ward off eye poke, eye poke is averted, but it hurts!*
*rolls backward and jumps limberly onto feet*
*makes the Matrix hand gesture, saying com'on grrl... (hiding the hurt of the averted eye poke)*

*bolts!*

The centepital force then makes up the entire inertial side and is =ma=mrw^2

How interesting. Where can I do more reading about it?

It is better to use horizontal (x) and vertical (y) components of forces than normal and tangential ones, as the resultant force is horizontal, and the only force we know -gravity- is vertical. You have to choose the positive direction both for x and y.

Hmm, you're saying it's better not to use my coordinates along the surface? It's just that I got used to it from statics as it sometimes helped simplify calculations.

N is the magnitude of the normal force, it is just N, no matter what mg is. What are its horizontal and vertical components in terms of the angle of the slope?
Fs is the static friction and its direction is so that it prevents the body from upward motion, according to the problem text, so the force of friction points downward.
What are its vertical and horizontal components?

When you have all these components of forces, collect the horizontal ones which result in the centripetal force, and collect all the vertical ones, which make zero.

Could you please show what you get?

Before I do that can you just tell me what coordinates should I use? Along the incline or standard one?

 

[ehild]

Try the standard one first, but do something, please. I have to leave in a hour.

ehild.