Dragging a Rod - Variation

Classical Mechanics Level 3

A uniform rod is lying tilted on a flat surface. The rod is touching the surface at its two ends, and it is supported there. The coefficient of friction is $\mu$ . The surface makes an angle of $\gamma$ with the horizontal. Initially, the rod makes an angle of $\alpha>0$ with the horizontal line drawn at its lower end, as shown.

Now, the lower end of the rod is dragged along the horizontal line in the same direction as the top is leaning to. There are two possible outcomes:

The rod may rotate upwards and eventually reach (and pass) a position perpendicular to the dragging line, as in this problem .
The rod starts to rotate downwards, eventually flipping over to the other side of the line.

What is $\alpha$ , in degrees, that separates the two outcomes, if $\mu=0.5$ and $\gamma = 15 ^{\circ}?$

The answer is 28.2.

1 solution

Laszlo Mihaly
May 24, 2018

We pick the flat surface as the basis of our reference frame: the $x$ axis is along the dragging line, the $y$ axis is in the plane perpendicular to the $x$ axis, and the $z$ axis perpendicular to the surface. The components of the weight vector will be $(0, -mg \sin \gamma, -mg \cos \gamma)$ . The $z$ component of the weight is balanced by the surface at the two ends of the rod, distributed evenly between the two support points. Therefore the normal force acting at the free end is $(1/2) mg \cos \gamma$ and the magnitude of the friction force is $(1/2) \mu mg \cos \gamma$ . The $y$ component of the force of gravity acts at the center of the rod, with magnitude $mg \sin \gamma$ .

If we are not at the limiting angle and we start to pull the rod, it will rotate either upwards or downwards. At the limiting angle the rod will not rotate at all, and therefore the top is moving parallel to the bottom. The friction force will point to the negative $x$ direction. If there is no rotation, the net torque acting on the rod must be zero. It is best to select the pulled end as the axis of rotation, and we get for the torque:

$(\frac{1}{2}\mu mg \cos \gamma) (\ell \sin \alpha) = (mg \sin \gamma )(\frac {\ell}{2} \cos \alpha)$

Where $\ell$ is the length of the rod. The solution is

$\tan \alpha = \frac {1}{\mu} \tan \gamma$

Notes:

It is easy to see that dragging at the critical angle is not a stable situation. Any deviation from the angle will lead to a torque that increases the deviation.
When setting up this problem one has to be careful not to make a surface too steep. If $\alpha> \tan^{-1}\mu$ , the whole set up is unstable.
Independent of the starting conditions, the rod will end up behind the pulled end, leaning downwards by the critical angle.

Is there any particular reason for the emphasis on the perpendicular position? Because the angle doesn't appear to stabilize at 90 degrees once it gets there.

Steven Chase - 3 years ago

This is only a refrerence to an existing problem. The final position is the same (described in the notes to my solution), independent of the initial conditions.

Laszlo Mihaly - 3 years ago

Dragging a Rod - Variation

The answer is 28.2.

1 solution

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