Is it possible for a vertical rod balancing on a table to lose contact by striking the top of the rod?

[Rob2024]

Homework Statement

An upright rod of length $l$ with negligible mass is initially at rest on a frictionless horizontal table. Two identical masses are attached to the top and the bottom of the rod. When the top of the rod is given a large horizontal velocity $v_0$, the bottom can lose contact with the table at the same moment.What's the minimum speed $v_0$ that allows this to happen?

Relevant Equations

$$F = ma$$

The question I have is that if this is even possible?

Assume it's possible, the CM acceleration is $g$ downward. Then the top end of the rod has an additional downward acceleration $a_c$ while the bottom end of the rod has an upward acceleration $a_c$. This will make the two ends of the rod having different acceleration in magnitude which contradicts with the given the rod is rigid. It suggests it's not possible for the system to lose contact with the table.

 

[haruspex]

Assume it's possible, the CM acceleration is $g$ downward. Then the top end of the rod has an additional downward acceleration $a_c$ while the bottom end of the rod has an upward acceleration $a_c$. This will make the two ends of the rod having different acceleration in magnitude which contradicts with the given the rod is rigid. It suggests it's not possible for the system to lose contact with the table.

If a rod spins about its midpoint, each end is accelerating towards the centre.

 

[TSny]

Assume it's possible, the CM acceleration is $g$ downward. Then the top end of the rod has an additional downward acceleration $a_c$ while the bottom end of the rod has an upward acceleration $a_c$. This will make the two ends of the rod having different acceleration in magnitude...

Yes

...which contradicts with the given the rod is rigid.

Rigidity imposes a condition on the velocities of two points of the rod, not the accelerations.

 

[jbriggs444]

This will make the two ends of the rod having different acceleration in magnitude which contradicts with the given the rod is rigid.

I just tossed a pen in the air, giving it an end over end spin. The laws of physics did not prevent it.

 

[Rob2024]

I get it now. My assumption about acceleration was incorrect.

 

[wrobel]

I would assume that the bottom of the rod can slide along a horizontal line without friction. Then I would answer the problem's question by considering a reaction of this ideal constraint.

 

[haruspex]

I would assume that the bottom of the rod can slide along a horizontal line without friction. Then I would answer the problem's question by considering a reaction of this ideal constraint.

It is stated as frictionless, but it makes no difference anyway.

 

[DEvens]

That's a pretty little problem. Yes, it is possible for the lower mass to lose contact with the table. The motion right after that is a lot more complicated to work out. But recall your formula for centripetal acceleration. And note that at the t=0 moment you have a lot of really cool simplification to your angles and velocities and such. And that the force a rigid massless rod applies has to be parallel to the rod.