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There is a floor that friction is proportional to its velocity (like $F=-kv$) and there is a box with its width as $l$ and its height as $h$. (you may assume that $l$ is longer than $h$). It is on the floor with the initial velocity $v$. Then, if $v$ is big, I think it would rotate. But I don't know how to analyze this situation as differential equation.

This figure shows the situation it rotate as the $\angle \theta$. But, as the initial time, $\theta$ is $0$ so thus I think its friction would not be exerted at the left-bottom corner. So it will be different with this figure.

If there is no floor, the rotate pivot is just the center of mass I think. But there is a floor. It makes me crazy.

My situation

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  • $\begingroup$ You have the normal force incorrect. (and therefore, the friction force is incorrect too). The normal force should ensure that the acceleration of the corner is zero in the y direction (it is at position $(-\cos(\theta+\theta_2)r+x,\sin(\theta+\theta_2)r+y)$, where $\theta_2$ is some constant angle made between the bottom of the rectangle and the line between the corner and the center, and r is the distance between the corner and the center). Solving this gives you three forces, and you will need to convert them to torque and force on the center of mass to get the final eqs. $\endgroup$ – user12029 Oct 5 '13 at 0:17
  • $\begingroup$ Related: physics.stackexchange.com/a/41908/392 $\endgroup$ – ja72 Apr 3 '14 at 15:10
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I'll make this easy for you. Always remember, torque is always calculated about the centre of mass of the object.When the object is lying on the floor, the normal reaction and weight pass through the centre of mass. So no torque from them. But the friction acts at a distance of h/2 from centre of mass. Distance is always calculated perpendicular to the direction of force called the lever arm. So a net torque kV x h/2 will be there. So yes the object will tend to rotate. Now as it starts rotating, the bottom left corner will become the pivot. The normal reaction will start acting at the pivot and your diagram is correct. Now the normal reaction also has a lever arm with the centre of mass, it will producetorqtorque but in the opposite direction. Torque is given by mg x l/2. If this torque is greater than the torquedue to friction, the object will not rotate. Comparing both torques you can easily see that the object will rotate if v > mgl/kh.

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  • $\begingroup$ Note that this site does have MathJax enabled, so that you can add easy-on-the-eyes mathematical formulae to your post. $\endgroup$ – Kyle Kanos Oct 14 '15 at 19:36
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Even with non-zero friction, the object would rotate. The object is not in equilibrium (even though the vertical forces are balanced, the velocity produces a torque, and so the net forces on the object are not zero). Friction is not enough to stop the rotation, because it is only a fraction of the normal force. F = kv, F = dp/dt (You would be better off using force = dp/dt, which goes to mdv/dt= kdv. And torque would simply be the cross product of the force and h, as h is perpendicular to the box, and the lever arm is perpendicular to the object being rotated. Hope this helps.

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    $\begingroup$ I just pushed my keyboard on my desk and it slid without rotating even after I let go. Why is that? $\endgroup$ – Brian Moths Oct 5 '13 at 1:46
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Velocity cannot produce torque. Only forces acting from a distance can produce torque.The friction acts on the entire base of the box to prevent its relative slipping. Here, the frictional force is proprotional to the velocity applied. Hence considering the horizontal forces $$F=-kv$$ $$ma=-kv$$ $$mdv/dt=-kv$$ $$mdv/v=-kdt$$ Integrate with proper limits to get the velocity at any time t.

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My analysis is that the object would not rotate at any velocity. My reasoning is that if the object is completely flat on the floor then there would be no net torque. Say r is the distance from the bottom left corner (as in your diagram) to the point at which the net frictional force acts (directly beneath center of mass, at a point where the object and floor are in contact). In this case r is parallel to the floor, and so is the net frictional force, f. Therefore: r x f = 0 (no frictional torque) The weight and normal force are both equal and opposite and act through the object's center of mass, therefore their torques would cancel each other out. This means that there is no net torque, regardless of the value of the frictional force. The object would not rotate.

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