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So, we can use the torque $$T = I\alpha$$to calculate the torque and thus the acceleration of friction force when the object is rolling without slipping. Then plug that in into $$F = ma$$ (As acceleration caused by torque is the acceleration of the object). If the Friction Force is responsible for the object's acceleration, why in the $F = ma$ part do we write that friction actually SLOWS IT DOWN. And assuming that it is a cylinder that is rolling, ($I = 0.5 mR^2$) if we have a force pushing that is equal to the friction force, shouldn't it itself cause the wheel to rotate?

edited: If i said something unclear then i am sorry. i asked: The static friction force has a limit, before that, the static friction should equal the force applied. It would be logical to assume that the body is a rigid body, and the force applied is the same in every point of it, so the friction should equal the force at any point in time with ANY FORCE APPLIED. if we think that it is rolling without slipping, a.k.a. friction force is less then maximum static friction force. thus friction is THE SAME AS THE FORCE APPLIED. Then according to F = ma the acceleration is zero. why am I wrong.

But according to $F = ma$ the acceleration is $0$. But if the friction is static, then it is either less than or equal to the force with which it is being pushed, so if there is no slipping, then the friction force is equal to the force with which the object is pushed, so acceleration should be zero. WHAT IS HAPPENING? HELP.

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It would be logical to assume that the body is a rigid body, and the force applied is the same in every point of it, so the friction should equal the force at any point in time with ANY FORCE APPLIED.

This is your mistake. There is no reason to assume that the static friction force magnitude is equal to the applied force magnitude in general. You are correct to think in terms of Newton's second law though. Applying a force $F_\text{app}$ from a distance $r$ above the center of the object of radius $R$ and in a direction perpendicular to the radius give us

Net force: $$F_\text{app}-f_s=ma$$

Net torque: $$rF_\text{app}+Rf_s=I\alpha$$

And then additionally imposing the rolling without slipping condition $a=R\alpha$ results in the correct relation between $f_s$ and $F_\text{app}$

$$f_s=\frac{I-mrR}{I+mR^2}F_\text{app}$$

and also gives us the acceleration of the object $$a=\frac{rR+R^2}{I+mR^2}F_\text{app}$$

So let's see what this tells us

1) The only way for $F_\text{app}=f_s$ is for $r=-R$. This corresponds to a sign change in the torque of the applied force, which corresponds to our force being applied below the center of the object instead of above, but keeping its same direction. However, also note that this causes $a=0$. Therefore, you can get what you were reasoning to, but this is not true in general.

So, for example, if you tied a string to the bottom of your cylinder and pulled horizontally on it, the static friction would balance out your applied force until slipping occurs. However, if you tied your string anywhere else, we would not get $F_\text{app}=f_s$, and you would get rolling without slipping (as long as you don't apply too much force).

2) Depending on how $I$ relates to $mrR$ the static friction force can act in either direction. In our case $I>mrR$ means the static friction force acts in the opposite direction of the applied force, and $I<mrR$ means that the static friction force acts with the applied force. Note that this is for when we apply the force above the center of the rolling object.

In the case where $I=mrR$ this just means that we will get rolling without slipping without the need of any static friction force. For example, you could get a hoop to roll without slipping on ice by just applying a force to the top of the hoop.


An additional point of intuition

If the Friction Force is responsible for the object's acceleration, why in the $F=ma$ part do we write that friction actually SLOWS IT DOWN.

By imposing rolling without slipping condition, we are essentially requiring a balance between translational and rotational motion in such a way that $x=R\theta$ , $v=R\omega$, and $a=R\alpha$ (really these are all saying the same thing). So, if there is no slipping, for a given applied force, we require the static friction force to "pick up the slack", so to speak. For example, if our applied force alone will cause the translation to be "faster than" the rotation, then we would need static friction to come in an slow down the translation and/or speed up the rotation. If what we need of static friction in order to do this is too much for static friction to handle, we will get slipping.

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  • $\begingroup$ Thank you very much, I think this explains everything quite well to me. But one last thing is there unclear to me...In order for it to be a rolling motion, the sum of forces at the bottom has to be zero and sum of velocities too, a.k.a. the bottom should be instantaneously at rest. There should be a force at top to keep it rolling or flipping so to speak. So this should contradict the statement that forces should not be equal at the bottom. Or am I just being a little too stupid...Sorry. $\endgroup$ – GameOver Apr 28 at 4:52
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  • Frictional force < maximum static friction: If the frictional force is less than the statice friction, then the cylinder will roll without slipping. I am assuming force is applied at the centre of the cylinder, so there will be a net torque due to frictional force, as $\tau=I\alpha$, there must be some finite angular acceleration. Now since there is rolling without slipping, $$a_{cm}=r\alpha$$ So there must be some linear acceleration. Using $F=ma$, $$F_{applied}-f_{friction}=ma_{cm}$$ $$F_{applied}=f_{friction}+ma_{cm}$$ The applied force can't be equal to friction.
  • Frictional force > maximum static friction: In this case, there will be rolling with slipping, so if $f_{friction}=\mu_{k}N$, N is the normal force (can be taken as $mg$ if on the horizontal surface), then $$\tau=\mu_{k}Nr=I\alpha$$ $$\alpha=\frac{\mu_{k}N}{I}$$ Using $F=ma$, $$F-\mu_{k}N=ma_{cm}$$ So the applied force can be equal to friction if the linear acceleration is zero, but it can still have angular acceleration.
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  • $\begingroup$ If i said something unclear then i am sorry. i asked: The static friction force has a limit, before that, the static friction should equal the force applied. It would be logical to assume that the body is a rigid body, and the force applied is the same in every point of it, so the friction should equal the force at any point in time with ANY FORCE APPLIED. if we think that it is rolling without slipping, a.k.a. friction force is less then maximum static friction force. thus friction is THE SAME AS THE FORCE APPLIED. Then according to F = ma the acceleration is zero. why am I wrong. $\endgroup$ – GameOver Apr 27 at 14:57
  • $\begingroup$ @GameOver The static friction force does not need to be equal to the applied force, and this answer shows why. In this case it actually will not be, in general. $\endgroup$ – BioPhysicist Apr 27 at 14:59
  • $\begingroup$ what is the intuition behind those? I understand the equation but i can't understand it practically. I don't understand why do we subtract the friction from the applied force if friction IS the acceleration. and then there must be a good way to calculate the friction. thank you by the way. $\endgroup$ – GameOver Apr 27 at 15:02
  • $\begingroup$ @GameOver See my answers on the related questions I just commented on your main post here $\endgroup$ – BioPhysicist Apr 27 at 15:04
  • $\begingroup$ @GameOver When you say "logical to assume that the body is a rigid body, and the force applied is the same in every point of it, so the friction should equal the force at any point in time with ANY FORCE APPLIED", this statement is not correct. This is correct when you have a point object. When you have rigid body you have to be careful about where the applied force is acted. If the applied force is at the point of friction force, then there will be zero acceleration and zero angular acceleration. $\endgroup$ – sslucifer Apr 27 at 16:08

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