Rotational mechanics question?

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Discussion Overview

The discussion revolves around a question related to rotational mechanics, specifically focusing on the forces acting on an object rolling down an inclined plane. Participants explore concepts of friction, energy conservation, and the dynamics of rolling motion, including the relationship between linear and angular acceleration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions why an object with large friction moves downward instead of upward, suggesting a conflict with common experience.
  • Another participant asserts that moving upward would violate the law of conservation of energy, as it would imply an increase in potential energy without external work.
  • There is a discussion about the nature of static friction, with some participants arguing that it can adjust to prevent slipping, rather than being fixed at a maximum value.
  • Participants propose calculations involving torque and moment of inertia, with varying interpretations of the role of friction in achieving pure rolling motion.
  • Some participants express confusion over the significance of the angle of inclination and its representation in equations, leading to further clarification requests.
  • There is a debate about the correct application of Newton's laws to derive the friction force and the conditions under which different equations apply.
  • One participant suggests that the friction force needed to prevent slipping depends on the rotational inertia of the object, while another emphasizes the importance of using consistent symbols for radius and radius of gyration.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions regarding friction and rolling motion. While some calculations are acknowledged as correct for specific cases, there is no overall consensus on the broader application of these principles or the interpretations of the variables involved.

Contextual Notes

Participants highlight limitations in their understanding of the significance of certain variables, such as the angle of inclination and the nature of rotational inertia. There is also mention of the need for clarity in variable representation, which may affect the conclusions drawn from the equations.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of physics, particularly those interested in rotational dynamics, energy conservation, and the mechanics of rolling objects on inclined planes.

vkash
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One more question on rotation.

first of all see attachment i want to say that if friction is too large then net force is in direction of inclined in upward direction but from common experience we can say that it will move in downward direction so where is it going wrong. why it moves in downward rather than upward.
 

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If it goes upward then you are breaking law of conservation of energy.You are increasing the energy(potential) of system by itself without external work being done.
 
vkash said:
first of all see attachment i want to say that if friction is too large then net force is in direction of inclined in upward direction but from common experience we can say that it will move in downward direction so where is it going wrong. why it moves in downward rather than upward.
Your error is in thinking that friction must equal μmgcosθ. But the static friction will be whatever it needs to be to prevent slipping, up to that maximum value.
 
Doc Al said:
Your error is in thinking that friction must equal μmgcosθ. But the static friction will be whatever it needs to be to prevent slipping, up to that maximum value.

I also thinking so.
frictional force will equal to mgsin(t). Net force in direction of inclined will zero and it will in pure rolling motion. Am i correct here? I am in doubt about this.
 
Last edited:
Hey, but what about energy conservation?
 
1994Bhaskar said:
Hey, but what about energy conservation?
bhaskar bhai. I have made a solution on the paper.Can you please check this.
Hope this is correct.through the figure.
torque around IAOR(instantaneous axis of rotation) is T.
moment of inertia about IAOR is Mk2. think that k is radius of gyration.Since friction is too large so it will static and enough to have pure rolling. a is acceleration linear acceleration of center of mass(of object on incline) and A is angular acceleration.
now calculation
T=IA=mgsin(t) R
A= [itex]\frac{gsin(t)}{2R}[/itex].
a=[itex]\frac{gsin(t)}{2}[/itex]
So here we can conclude that frictional force is 0.5*m*g*sin(t).tell me if i a wrong somewhere.
 
vkash said:
I also thinking so.
frictional force will equal to mgsin(t). Net force in direction of inclined will zero and it will in pure rolling motion. Am i correct here?
No, you are not correct. You can figure out the friction force by applying Newton's law to both translation and rotation.

If you call the friction force F, then the net force will be mgsinθ - F. That same force F will also produce a torque on the sphere. By applying Newton's law and the condition for rolling without slipping you can solve for F.
 
vkash said:
bhaskar bhai. I have made a solution on the paper.Can you please check this.
Hope this is correct.


through the figure.
torque around IAOR(instantaneous axis of rotation) is T.
moment of inertia about IAOR is Mk2. think that k is radius of gyration.Since friction is too large so it will static and enough to have pure rolling. a is acceleration linear acceleration of center of mass(of object on incline) and A is angular acceleration.
now calculation
T=IA=mgsin(t) R
A= [itex]\frac{gsin(t)}{2R}[/itex].
a=[itex]\frac{gsin(t)}{2}[/itex]
So here we can conclude that frictional force is 0.5*m*g*sin(t).


tell me if i a wrong somewhere.
I don't quite follow what you're doing. See my last post for an outline of how to solve for the friction force.
 
Well I didn't understand the significance of 't'. It's okay that mu(coefficient of friction)*g*sin(x) is limiting case and that's what you have to consider. If you can explain me the physical significance of 't'(that's your mistake) then it's okay what you have done(For any middle case).
 
  • #10
Doc Al said:
I don't quite follow what you're doing. See my last post for an outline of how to solve for the friction force.

friend I am doing the same thing but about the instantaneous of rotation. You may not understanding my way because i skip many step.

If I do in the way you say then.
torque about the center of
Fr=mr2a
F=mra
F=mA.
Now
mgsin(t)-F=mA
since mA=F
so mgsin(t)-F=F
2F=m*g*sin(t)
so F=0.5*m*g*sin(t).
t is angle of inclination of incline, a is angular acceleration, A is linear acceleration.

Am i correct??
 
  • #11
Sorry didn' read it completely
 
  • #12
1994Bhaskar said:
Well I didn't understand the significance of 't'. It's okay that mu(coefficient of friction)*g*sin(x) is limiting case and that's what you have to consider. If you can explain me the physical significance of 't'(that's your mistake) then it's okay what you have done(For any middle case).

t is angle of inclination of inclined plane. In figure it is x. that's Variable mismanagement.
 
  • #13
So am i correct now?
 
  • #14
mA=F is final limiting case.I thought you wanted to calculate for case in between
 
  • #15
vkash said:
If I do in the way you say then.
torque about the center of
Fr=mr2a
So: F = friction; I = mr2 (a cylindrical shell?); a = angular acceleration
F=mra
F=mA.
Now
mgsin(t)-F=mA
since mA=F
so mgsin(t)-F=F
2F=m*g*sin(t)
so F=0.5*m*g*sin(t).
t is angle of inclination of incline, a is angular acceleration, A is linear acceleration.

Am i correct??
Realize that you are solving for a special case where I = mr2, a cylindrical shell. But given that, you are correct.
 
  • #16
Doc Al said:
So: F = friction; I = mr2 (a cylindrical shell?); a = angular acceleration

Realize that you are solving for a special case where I = mr2, a cylindrical shell. But given that, you are correct.

If r is radius of gyration. then will it correct?

1994bhaskar said:
mA=F is final limiting case.I thought you wanted to calculate for case in between
what you mean. I did not understand. can you please explain.
 
  • #17
vkash said:
If r is radius of gyration. then will it correct?
No. The friction force needed to prevent slipping will depend on the rotational inertia.

Don't use the same symbol for radius and radius of gyration.
 
  • #18
Doc Al said:
No. The friction force needed to prevent slipping will depend on the rotational inertia.

Don't use the same symbol for radius and radius of gyration.

I can't finally. I have made my tries. I am still thinking that I am correct. This time i will say dear stop this chat and answer me.:cry:(friction and acceleration of object):cry:
 
  • #19
vkash said:
I can't finally. I have made my tries. I am still thinking that I am correct. This time i will say dear stop this chat and answer me.
:confused: Answer what? Do you not see where you assumed that I = mr2? If you want to be more general, use I = mk2. But then your answer will be in terms of k and r, which will no longer cancel.

If you pick the object you want to study, then you can compute its acceleration and the required friction.
 
  • #20
Doc Al said:
:confused: Answer what? Do you not see where you assumed that I = mr2? If you want to be more general, use I = mk2. But then your answer will be in terms of k and r, which will no longer cancel.

If you pick the object you want to study, then you can compute its acceleration and the required friction.

take it a ring of radius R and solve. find friction and acceleration. If you think there is still lag of any variable than assume that but just do it.
 
  • #21
vkash said:
take it a ring of radius R and solve. find friction and acceleration. If you think there is still lag of any variable than assume that but just do it.
See my post #15. You've already solved for that special case and I agree with your answer.
 
  • #22
Doc Al said:
See my post #15. You've already solved for that special case and I agree with your answer.

So finally close this thread.
thanks for helping. I have one more doubt in rotational mechanics. Can you please hep me in this question also.

Once again thanks for answering.
 
  • #23
vkash said:
I have one more doubt in rotational mechanics. Can you please hep me in this question also.
You're making a similar error in that thread, as vela has already pointed out. Given what's been discussed in this thread, solve for the translational acceleration of a disk.
 

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