Rotational mechanics question?

[vkash]

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.

 

[1994Bhaskar]

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.

 

[Doc Al]

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.

 

[vkash]

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.

 

[1994Bhaskar]

Hey, but what about energy conservation?

 

[vkash]

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= $\frac{gsin(t)}{2R}$.
a=$\frac{gsin(t)}{2}$
So here we can conclude that frictional force is 0.5*m*g*sin(t).tell me if i a wrong somewhere.

 

[Doc Al]

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.

 

[Doc Al]

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= $\frac{gsin(t)}{2R}$.
a=$\frac{gsin(t)}{2}$
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.

 

[1994Bhaskar]

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).

 

[vkash]

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??

 

[1994Bhaskar]

Sorry didn' read it completely

 

[vkash]

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.

 

[vkash]

So am i correct now?

 

[1994Bhaskar]

mA=F is final limiting case.I thought you wanted to calculate for case in between

 

[Doc Al]

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.

 

[vkash]

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?

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.

 

[Doc Al]

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.

 

[vkash]

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.(friction and acceleration of object)

 

[Doc Al]

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.

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.

 

[vkash]

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.

 

[Doc Al]

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.

 

[vkash]

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.

 

[Doc Al]

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.