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This is a question I was trying to solve: A uniform disk of mass 1 kg and radius 1 m is kept on a rough horizontal surface. Two forces of magnitudes 2 N and 4 N have been applied on the disk as shown in the figure. If there is no slipping then the linear acceleration of the centre of mass of the disk is? enter image description here

The solution offered here says that the torque about Point C is 4R- 2(2R)= zero, so the Net force is zero and hence linear acceleration of centre of mass is zero.

I don't understand how the net force becomes zero, just because the torque about point C is zero.

Does torque=0 always mean net force=0?

For the condition that slipping does not occur, the condition that I learnt of required 1) point C to be at rest wrt the surface and 2) if linear acceleration present, linear acceleration= (angular acceleration)(radius) [a=αR]. So how does this condition of torque about point C satisfy the body not slipping?

Any help is appreciated.

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This is an example of a trick that often comes in handy when analyzing the motion of rigid bodies: if there's a single unknown force, compute the torques about the point where that force acts. Regardless of the magnitude or direction of the unknown force, it will contribute nothing to the torque, because the moment arm is zero. So, you can calculate the instantaneous angular acceleration of the body about the chosen point using only the known forces.

In this case, the forces have been chosen so that the torque about point C is zero: the two applied forces create equal and opposite torques, and the gravitational force acts on a line through point C, so contributes no torque. Thus, there's no acceleration about point C, and if the disk is initially at rest, it remains so.

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  • $\begingroup$ Thanks for the trick! How did you conclude that because torque is zero, the linear acceleration is zero as well? We don't know if the friction is equal to 2 N. $\endgroup$ – laksheya Sep 25 '19 at 15:14
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    $\begingroup$ That's right! We don't know that a priori. However, it must be so, because we found on the basis of torques that the thing doesn't move. If there is no slipping, and no rotation (since torque is zero), that is sufficient to establish that the disk does not start to move. If it doesn't move, its linear acceleration is zero. If either A) it was allowed to slip, or B) an applied force was altered so that torque about C was no longer zero, then linear acceleration would be nonzero. $\endgroup$ – Ben51 Sep 25 '19 at 15:18
  • $\begingroup$ Thank you so much! This helped me a lot! :) $\endgroup$ – laksheya Sep 25 '19 at 15:23
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I don't understand how the net force becomes zero, just because the torque about point C is zero.

Since there is no slipping, this means that the maximum static friction force at C has not been exceeded. That further means that there is a static friction force of 2 N acting to the left which equals the applied horizontal force of 2 N to the right. Or, the sum of the horizontal forces equals zero. The normal reaction at the surface acting up at C equals the weight of the disc acting down, so the sum of the vertical forces equals zero. Since there is no net force acting on the disc, it will not accelerate.

Does torque=0 always mean net force=0?

If you mean sum of the torques (sum of the moments), then the answer is no. In statics the preferred term is moment, but it is equivalent to torque since the moment about a point it is the product of a force times the perpendicular distance between the point and the force.

Similarly, just because the sum of the forces equals zero doesn't necessarily mean the sum of the moments is zero. The sum of moments and forces are two independent criteria for equilibrium. Moments address rotational motion. Forces address translational motion.

For the example given, to complete the conditions of equilibrium, there must also be no rotation. That means the sum of the moments (torques) about any point must equal zero. Taking the moments about point C you get 4R-2(2R)=0 per the solution. The static friction force acts through point C so it contributes no moment (moment arm=0) about point C.

So how does this condition of torque about point C satisfy the body not slipping?

It is not the condition of torque about point C that satisfies not slipping. It is the condition that the applied force acting to the right (2 N) does not exceed the maximum static friction force to the left which is $μ_{s}mg$ where $μ_s$ is the coefficient of static friction. At that point, slipping is impending. Below that level the static friction force acting to the left simply matches the force acting to the right so that there is no net horizontal force and thus no slipping.

The condition of sum of the moments (torques) addresses rotation. You can have pure rotation, without slipping. When the sum is zero, there is no rotation.

Hope this was not too long and that it was of some help.

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  • $\begingroup$ Torque is zero, you have mentioned product of distance into r either f=0 and or r=0 for a particular force producing torque, but when we discuss it about sum of all the torque about is zero there can also be the case where all particles are at point of reference about which the torque is taken @bob $\endgroup$ – Yuvraj Singh... Sep 25 '19 at 13:33
  • $\begingroup$ Thank you for your answer. What I understood from your answer is that for the condition of not slipping, the net force should be zero. Isn't that a false statement? The body can still be accelerated and not slip because the tangential acceleration αR balances the linear acceleration a at the point C. Please correct me if I'm wrong. $\endgroup$ – laksheya Sep 25 '19 at 13:34
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    $\begingroup$ @laksheya That is not what I am saying. I am saying there is no slippage because the static friction force to the left equals the applied 2 N force to the right. In this problem there is neither a net torque nor a net force on the disc and therefore it neither rotates nor accelerates. If it were not for the torque due to the downward 4 N force, there would be a net clockwise torque. Then the static friction force would enable the center of mass to accelerate without slippage. $\endgroup$ – Bob D Sep 25 '19 at 15:50
  • $\begingroup$ @yuvrajsingh Not quite sure what you are getting at. In this problem it is clear that both the sum of the moments and sum of the forces are zero and therefore the disc neither rotates nor accelerates. If it were not for the counter-clockwise torque due to the 4 N force, the wheel would rotate and accelerate without slippage. $\endgroup$ – Bob D Sep 25 '19 at 15:53
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    $\begingroup$ @laksheya Yeah, I thought that might have been the case. In retrospect I should have discussed the sum of the torques being zero first preventing rotation and then discuss static friction as preventing slipping. $\endgroup$ – Bob D Sep 25 '19 at 16:06

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