What is the equation of motion to determine when a rod will slip off a table?

[schuksj]

Hi. My question is this. A uniform rod of length 2L is held resting on a horizonal table with length L+a projecting over the edge. If the support is removed, show tthat the rod willb egin to slide over hte ege when it has turned through an angle tan^-1(mu*L^2/(L^2+ 9a^2)). mu is the coeffiecent of static friction. I am having trouble finding the equation of motion to start this problem.


m*a=mgsin(theta)-mu*m*g*cos(theta) I think is the right way to start since
force of friction=mu*m*g*cos(theta). Now how do I go from here to get the angle it slips off? And what should the angle of rotation be?

 

[Andrew Mason]

Hi. My question is this. A uniform rod of length 2L is held resting on a horizonal table with length L+a projecting over the edge. If the support is removed, show tthat the rod willb egin to slide over hte ege when it has turned through an angle tan^-1(mu*L^2/(L^2+ 9a^2)). mu is the coeffiecent of static friction. I am having trouble finding the equation of motion to start this problem.


m*a=mgsin(theta)-mu*m*g*cos(theta) I think is the right way to start since
force of friction=mu*m*g*cos(theta). Now how do I go from here to get the angle it slips off? And what should the angle of rotation be?

Are you turning the table up?

Start by analysing the forces on the rod under static conditions:

$$F_{gravity} + F_{normal} + F_{friction} = 0$$

The normal force is the trickiest because there is a torque about the edge of the table that has the effect of reducing the normal force on the part that is in contact with the table.

AM

 

[OlderDan]

Hi. My question is this. A uniform rod of length 2L is held resting on a horizonal table with length L+a projecting over the edge. If the support is removed, show tthat the rod willb egin to slide over hte ege when it has turned through an angle tan^-1(mu*L^2/(L^2+ 9a^2)). mu is the coeffiecent of static friction. I am having trouble finding the equation of motion to start this problem.


m*a=mgsin(theta)-mu*m*g*cos(theta) I think is the right way to start since
force of friction=mu*m*g*cos(theta). Now how do I go from here to get the angle it slips off? And what should the angle of rotation be?

The force of friction is not mu*m*g*cos(theta). I took the normal force to be always perpendicular to the rod and the frictional force parallel to the rod. The frictional force is set equal to the component of gravity parallel to the rod [mgsin(theta)] plus the centripetal force required to keep the rod moving on a circular path. The rod slips when the frictional force reaches mu*N. The normal force is found from torque and angular acceleration considerations and finding the linear acceleration of the CM as the ratio [mgcos(theta) - N]/m. I also used conservation of energy before slipping. I found it convenient to work in terms of I_CM and I about the pivot point (I_CM + ma^2; parallel axis theorem) until the end.