Speed for loss of of contact between a dumbbell and a table

[A13235378]

Homework Statement

Consider a dumbbell formed by two equal balls and a rod of negligible mass and length l, positioned vertically on a flat and horizontal table. A horizontal and instantaneous speed is communicated to the bottom ball. Determine the highest value of this speed so that the bottom ball does not lose contact with the table.

Relevant Equations

Energy conservation
T = I . a , T = resulting torque , I = inertia momentum, a =angular acceleration

Attempted solution:

Consider the instant when the normal force of the lower ball is zero. Conserving energy:

$$\frac{mv^2}{2}+mgh_1=\frac{mv_1^2}{2}+\frac{mv_2^2}{2} + mgh_2$$

Applying the resulting torque to the upper ball where the rotation point is the lower ball.

$$T=I.a = ml^2.\frac{v^2}{l}$$

From that, I crashed because there are many parameters. Furthermore, I cannot draw any conclusions and information at the moment when the normal force becomes zero

Can anybody help me?
I just need you to have a little patience and understanding, because I'm still a very beginner in this dynamic part of the rotation

 

[PeroK]

I think it would be easier to consider the system in a frame moving with speed $v/2$. That would give you a nice symmetry to work with. Then focus on total KE of the system as a function of the angle of the bar.

 

[A13235378]

I think it would be easier to consider the system in a frame moving with speed $v/2$. That would give you a nice symmetry to work with. Then focus on total KE of the system as a function of the angle of the bar.

but at what point will the normal force be zero?

 

[PeroK]

but at what point will the normal force be zero?

Isn't that what you are trying to calculate?

 

[onatirec]

The normal force is zero when the centripetal acceleration = acceleration of gravity. All of the kinetic energy is from the instantaneous velocity of the bottom ball. This energy should cause the system to rotate about it's center of mass. Already, we can equate, in terms of energy, the linear velocity of the single mass with the angular velocity of the system. This angular velocity effects a centripetal acceleration on both mass which, when larger than gravity, counteracts the normal force.

 

[Lnewqban]

The instantaneous center of rotation should be the center of mass of the dumbbell, which should be moving vertically in free fall.

If initial horizontal velocity is too high, the lower mass will separate from the table.
If initial horizontal velocity is too low, the lower mass will put some force on the table and will interfere with the vertical free fall of the CM.

 

[haruspex]

The normal force is zero when the centripetal acceleration = acceleration of gravity.

Centripetal acceleration of what? One ball or the dumbbell as a whole? And in what frame of reference?
Centripetal acceleration, by definition, is that component of overall acceleration which is normal to the velocity. In the frame @PeroK proposes, both linear velocity and acceleration of the mass centre are vertical, so no centripetal acceleration.

 

[haruspex]

The instantaneous center of rotation should be the center of mass of the dumbbell, which should be moving vertically in free fall.

You cannot have it both ways. If the mass centre is moving in your reference frame then it is not the instantaneous centre of rotation: https://en.m.wikipedia.org/wiki/Instant_centre_of_rotation. “the point fixed to a body undergoing planar movement that has zero velocity at a particular instant of time."

 

[onatirec]

Centripetal acceleration of what? One ball or the dumbbell as a whole?

Each ball will experience a centripetal force, since they are attached and the system must rotate from the instantaneous velocity of the lower ball (which is normal to the bar attaching the two balls).

In the frame @PeroK proposes, both linear velocity and acceleration of the mass centre are vertical, so no centripetal acceleration.

Unless PeroK is talking about a rotating reference frame, I fail to see how the two mass system would not be rotating and not be producing centripetal forces on each of the two masses. What else would cause the bottom ball to lift off of the table except the centripetal force of the rotating system?

 

[haruspex]

Each ball will experience a centripetal force

Ok, so you were talking about centripetal acceleration of an individual ball. That wasn't clear. But neither do I see how it is helpful. We are not concerned with e.g. the tension in the rod.

I fail to see how the two mass system would not be rotating

I didn't say it wasn't. I wrote that in @PeroK's frame of reference the mass centre of the dumbbell is not undergoing centripetal acceleration.

 

[haruspex]

but at what point will the normal force be zero?

Are you still working on this? @PeroK gave good advice in post #2: work in the inertial frame that moves horizontally with a velocity matching that of the horizontal component of the dumbbell's mass centre. In this frame, the mass centre only moves vertically.
Write down the energy conservation equation in terms of the angle of the bar and its derivative.
When/if the bar becomes airborne, what can you say about the behaviour of the angular velocity?