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Let us take a cylinder on a rough horizontal surface. The cylinder undergoes pure rolling A vertical downward force acts on the cylinder to produce a clockwise rotation. The point at which it acts can be any point on the cylinder provided it produces a clockwise rotation.

Now my goal was to find the direction of friction. I initially assumed friction acts leftward to produce a clockwise torque. But here's the issue. With a clockwise torque center of mass of the cylinder moves right. But there is no force acting on the cylinder in rightward direction. So such a situation is impossible. Thus I concluded that friction has to act right ward to produce a counterclockwise torque against the vertical force.

Is this line of reasoning correct? If not please explain why?

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If the cylinder is not accelerating and the rolling no slip condition $v_\textrm{COM} = R \omega$ is satisfied then the frictional force between the cylinder and the ground is zero.
The horizontal force on the cylinder is zero as it is not accelerating in that direction.
The weight of the cylinder is equal an opposite to the normal reaction pn the cylinder due to the ground.

If $v_\textrm{COM} > R \omega$ (spinning too slowly) then the translational velocity of the centre of mass needs to decrease and/or the rotational angular velocity of the cylinder needs to increase to achieve the rolling no slip condition.
In this case the direction of the frictional force will be in the opposite direction to the translational velocity of the centre of mass.

If $v_\textrm{COM} < R \omega$ (spinning too rapidly) then the translational velocity of the centre of mass needs to increase and/or the rotational angular velocity of the cylinder needs to decrease to achieve the rolling no slip condition.
In this case the direction of the frictional force will be in the same direction to the translational velocity of the centre of mass.

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  • $\begingroup$ Let me make your 10k rep, +1 ;)) $\endgroup$
    – user36790
    Jul 12, 2016 at 9:51
  • $\begingroup$ Thank you but I hope it was properly deserved. :-) $\endgroup$
    – Farcher
    Jul 12, 2016 at 11:30
  • $\begingroup$ Of course, you deserved that ;D $\endgroup$
    – user36790
    Jul 12, 2016 at 11:47
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You can easily find direction of friction force by drawing free body diagram for the cylinder. FBD of the cylinder is shown below (Note that I have drawn diagrams for counter clockwise rotation)

enter image description here

As you see in figure above, friction force $f$ opposes to rotation of the cylinder.

Equations of cylinder motion are as below $$f=ma_G\;\tag 1$$ $$N=mg+F\;\tag 2$$ $$Fd-fR=I_G\alpha\;\tag 3$$ $$a_G=R\alpha\;\tag 4$$ Equation $(4)$ is non-slipping condition.

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Your reasoning is completely correct.

Here is an alternative line of reasoning. I usually find it simpler.

  1. Cause of motion is clockwise torque. So friction will apply anticlockwise torque to oppose it.
  2. For anti-clockwise torque, friction has to act along right.
  3. Side effect of this friction is acceleration towards right.
  4. Right acceleration matches clockwise $\alpha$. Everyone is happy!
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