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

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SUMMARY

The discussion centers on calculating the maximum distance from the rotation axis at which a 6 kg object can be placed on a 30-degree inclined surface without sliding due to centripetal acceleration. The static friction coefficient is given as 0.4, and the period of rotation is 3 seconds. Participants emphasize the importance of understanding centripetal acceleration and the forces acting on the object, including the normal force and friction, to derive the necessary equations for solving the problem.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula: ac = v²/R
  • Knowledge of static friction and its coefficient
  • Familiarity with free body diagrams (FBD) and force analysis
  • Basic principles of rotational motion and angular velocity calculations
NEXT STEPS
  • Learn how to derive angular velocity from the period of rotation using the formula: ω = 2π/T
  • Study the relationship between centripetal acceleration and static friction in inclined planes
  • Explore the application of Newton's second law in rotational dynamics
  • Practice solving problems involving multiple equations and unknowns in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for problem-solving strategies in dynamics.

  • #31
Femme_physics said:
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! :smile:



Femme_physics said:
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)*


Femme_physics said:
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?
Femme_physics said:
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".
 
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  • #32
Femme_physics said:
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.

Femme_physics said:
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
 
  • #33
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
 
  • #34
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)*
:eek:
*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?
 
  • #35
Try the standard one first, but do something, please. I have to leave in a hour.

ehild.
 
  • #37
Good morning again Fp! :smile:


There you go! Very good!

Uh, one thing, doesn't Fs have an y component?
Add that, and go grrl, go! :wink:
 
  • #38
Good morning ILS :)And damn it I knew I forgot something. I got used to the inclined coordinates. Add Fs sin(30) in sum of all forces on y.
 
  • #39
Femme_physics said:
Good morning ILS :)


And damn it I knew I forgot something. I got used to the inclined coordinates. Add Fs sin(30) in sum of all forces on y.

That's it!
Now do these 2 equations scare you?
 
  • #40
Well, you know how the maximum of Fs is related to N. Use it in the equation to eliminate Fs. ehild
 
  • #42
I like the flowers and the butterflies! :smile:
How does it feel?

Uhh, except that I'm afraid I was careless with my last hint. Sorry! :redface:

Can you check if the y component of Fs is up or down?
That is, can you check the sign of "Fs sin 30"?EDIT: Oh, and I seem to be missing a square around the omega...
 
  • #43
Can you check if the y component of Fs is up or down?
That is, can you check the sign of "Fs sin 30"?

But it can't be wrong, I already drew flowers and butterflies. It can't. Flowers and butterflies means I'm right. Sorry, it's a cosmological fact. Whatever you think about mechanics is wrong because of that. Okay, fine, I'll fix the sign and get the right figure, but that's it basically, that is the answer right? I'll have to do some reverse engineering after I'm done to better follow what you've been trying to tell me.
 
  • #44
Femme_physics said:
But it can't be wrong, I already drew flowers and butterflies. It can't. Flowers and butterflies means I'm right. Sorry, it's a cosmological fact. Whatever you think about mechanics is wrong because of that.


Okay, fine, I'll fix the sign and get the right figure, but that's it basically, that is the answer right? I'll have to do some reverse engineering after I'm done to better follow what you've been trying to tell me.

Yep! That's it! :smile:

Oh, and take note of the missing square around the omega.
 
  • #45
Femme_physics said:
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?

Well you do consider all forces that act on the block (weight,normal,friction). But some of those forces do not have components in the radial direction such as weight. If you were to have no friction than the centripetal force is supplied purely by the component of the normal force that is in the radial direction.

The moment of inertial does not come into play unless the block is accelerating in a circle. Because just like F=ma, M=I(alpha) where 'M' is the net moment or torque, 'I' the moment of inertia, and 'alpha' the angular acceleration. Since you have no angular acceleration in this problem there is no torque applied to the block and no need to worry about the moment of inertia...
 
  • #46
Will this calculation FINALLY lead to my answer?

http://img718.imageshack.us/img718/9376/fxfxfxfxfxfxfxfx.jpg
Well you do consider all forces that act on the block (weight,normal,friction). But some of those forces do not have components in the radial direction such as weight. If you were to have no friction than the centripetal force is supplied purely by the component of the normal force that is in the radial direction.

The moment of inertial does not come into play unless the block is accelerating in a circle. Because just like F=ma, M=I(alpha) where 'M' is the net moment or torque, 'I' the moment of inertia, and 'alpha' the angular acceleration. Since you have no angular acceleration in this problem there is no torque applied to the block and no need to worry about the moment of inertia...

I think I somewhat understand you, I'd still love for a good source about centripetal force, the real forces that act, and how to get an understanding of it! I might have to make a different topic before it sinks in. I appreciate the feedback. :)
 
Last edited by a moderator:
  • #48
Why but of course I did!

*sneaky shifty look*

:wink:

Thank you ILS, and thank you everyone for helping me through this! :)
 
  • #49
Femme_physics said:
Why but of course I did!

*sneaky shifty look*

:wink:

Thank you ILS, and thank you everyone for helping me through this! :)
Hey! Not so fast! :smile:

Do you have the answer for R?
 
  • #51
Femme_physics said:
Sorry for the delay, but I always get the job done! ...eventually. :)

Is this the final equation I need to solve, then?

Yep! :smile:
 
  • #52
Score! :) Thank you ILS, Everyone. Long overdue!
 
  • #53
Femme_physics said:
Score! :) Thank you ILS, Everyone. Long overdue!

Hey! I want to see more flowers and butterflies (or some other artistic expression)! :smile:
 
  • #54
I would like to see solution for the minimal distance so as the object do not move downward.:smile:

ehild
 

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