What instant the force in the dipstick iz zero

[orlan2r]

We have the structure show in figure (ball mass m joined a dipstick mass is negligible). If the structure start to move of the position showed, find the angle between the dipstick and x-axis when the force in the dipstick es null (don't exists friction).

I've analized this problem but I arrived to a contradiction.

PD: I'm sorry for grammatical errors (I'm from Peru).

 

[tiny-tim]

welcome to pf!

hi orlan2r! welcome to pf!

show us your full calculations, and then we can see what went wrong, and we'll know how to help!

oh, and it's …

We have the structure shown in the figure (ball of mass m joined to a rod whose mass is negligible). If the structure starts to move from the position shown, find the angle between the rod and the x-axis when the force in the rod is zero (there is no friction).

I've analyzed this problem but I arrived at a contradiction

PS: I'm sorry for grammatical errors (I'm from Peru).​

(and the subject is mechanics, not mechanic )

are you a llama? I'm a goldfish, and I'm from london.

 

[rcgldr]

If the supporting "dipstick" has no mass, then only the mass of the ball matters. The dipstick has no mass, it has no linear or angular momentum. If the "smooth" surface is friction free, it takes zero force to move and/or rotate the dipstick, and I don't see how the dipstick at any angle other than vertical would offer any resitance to a falling ball, so I assume that the smooth surface does have a component of friction that is constant regardless of the speed of the bottom end of the dipstick. Separating the forces into vertical and horizontal components would help. The vertical components are constant, gravity pulls down on the ball, the surface pushes up with equal and opposing force.

The trick of this problem is to realize under what circumstance will the horizontal component of forces be zero. You've already stated that this occurs when the friction force is zero. Under what circumstance is the friction force between an object and a horizontal surface zero?

 

[tiny-tim]

If the supporting "dipstick" has no mass, then … I don't see how the dipstick at any angle other than vertical would offer any resitance to a falling ball.

uhh?

the "dipstick" is rigid …

that's a perfectly legitimate physical constraint …

the fact that its mass is zero (or negligible) just simplifies the maths.

 

[rcgldr]

If the "smooth" surface is friction free, it takes zero force to move and/or rotate the dipstick, and I don't see how the dipstick at any angle other than vertical would offer any resitance to a falling ball

confused ...

I editted my previous post. The issue I had was that "smooth surface" could imply friction free, in which case the friction is always zero, and there's never any force in the dipstick unless it's vertical. The problem makes sense only if there is friction between the dipstick and smooth surface.

 

[tiny-tim]

The issue I had was that … the friction is always zero, and there's never any force in the dipstick unless it's vertical.

No … the "dipstick" is rigid, and there will generally be a vertical reaction force from the surface, whatever the angle of the "disptick" itself.

 

[rcgldr]

The issue I had was that "smooth surface" could imply friction free, in which case the friction is always zero, and there's never any force in the dipstick unless it's vertical.

No … the "dipstick" is rigid, and there will generally be a vertical reaction force from the surface, whatever the angle of the "disptick" itself.

If there is no friction, there can be no horizontal forces involved. The rate at which the ball accelerates downwards is related to the forces and/or torques divided by the linear and/or angular moments of inertia. Since the dipstick has no mass, in the case of a frictionless surface, the dipstick offers no resistance to the free falling ball. It takes zero force and/or torque to move the dipstick out of the way as the ball falls.

The problem only makes sense if there is friction. It's complicated because the bottom of the dipstick doesn't move linearly until the horizontal component of force exceeds the force provided from the static coefficient of friction, which can be assumed to be the same as the dynamic coefficient of friction for this problem.

 

[orlan2r]

Thanks a lot tiny-tim and Jeff Reid for answer my post.

We consider the dipstick is a rigid body where their mass is negligible.

The common sense say to me the ball should fall vertically because over the system "ball + dipstick" only acting vertical forces (surface is friction free).

But for other side at beginning the force of reaction exerted for the ground over the dipstick is equal to "mg" and, newly, the common sense say to me that this force of reaction no will take a null value instantly.

The most probably is the value of this force decreases rapidly until this take a null value.

And if this situation is true, we could find the position in which the dipstick is separated from the ground.

This is the contradiction I've mentioned at beginning, but now, then to ear several commennt I think to have arrived to one conclusion: The force of reaction will take a null value just instantly.

Thanks again

 

[rcgldr]

(surface is friction free). ... force of reaction no will take a null value instantly. The most probably is the value of this force decreases rapidly until this take a null value.

One more possible issue.

The rod (dipstick) has no mass.
The surface is friction free.

I made a probably bad assumption that the rod had a frictionless hinged interface with the ball, in which case the rod is essentially non existant.

If the rod is rigidly attached to the ball, such as inserted into the ball, then the angular momentum and angular kinetic energy of the ball become factors. Perhaps this is the problem you're trying to solve?

 

[tiny-tim]

If the rod is rigidly attached to the ball, such as inserted into the ball, then the angular momentum and angular kinetic energy of the ball become factors. Perhaps this is the problem you're trying to solve?

Yes, the original diagram clearly shows a ball rather than a point mass, so the ball will rotate, and the motion will depend on the radius of the ball.

 

[rcgldr]

Yes, the original diagram clearly shows a ball rather than a point mass, so the ball will rotate, and the motion will depend on the radius of the ball.

Sorry for the brain fade, I'm a bit tired and out of it today, and my brain isn't fully functional. The problem statement just wasn't sinking into my thought process properly.