Tipping cylinder tube with balls in it

[Samuelriesterer]

Problem Statement

Two identical spheres of radius r are placed inside a cylinder of radius R as shown in the diagram below. You are given that R/2 < r < R and that each ball has a weight W. All surfaces are smooth. Show that there is a minimum mass, m, of the cylinder which will allow it to remain upright (for smaller cylinder masses, the cylinder tips over. Show that this mass is given by
m = (2W/g)(1-r/R).

Relative equations:

t = Fr sin theta
I = integral(r^2) dm

Work so far:

See attached picture and document. I have calculated d, the distance from the point of axis to the point of force as:

d = a + R = 2*sqrt(-r(r-2R)) + R

I think I am supposed to find the torque needed to tip the cylinder so I need to find the moment of inertia. But I am really unsure how to begin this problem.

Any help is appreciated.

Thanks!

 

[haruspex]

Two identical spheres of radius r are placed inside a cylinder of radius R
See attached picture and document. I have calculated d, the distance from the point of axis to the point of force as:

d = a + R = 2*sqrt(-r(r-2R)) + R

That looks right, except that you have swapped R and r from the problem statement.
Next is to find the normal forces. Draw FBDs for the spheres.
Don't forget the force of the lower sphere against the cylinder.

 

[Samuelriesterer]

Attached are my FBD and forces sums. I am unsure how to work the forces in this problem.

 

[haruspex]

Attached are my FBD and forces sums. I am unsure how to work the forces in this problem.

The idea is to have Fx and Fy equations for each sphere separately. That should tell what all the normal forces are.

 

[Samuelriesterer]

What am I solving for? There are many normal forces.

F_net m1x = Fnwl - Fnb sin o
F_net m1y = Fnt - Mbg -Fnb cos o

F_net m2x = Fnb sin o - Fnwr
F_net m2y = Fnb cos o - Mbg

F_net tx = Fnwr - Fnwl
F_net ty = Fnt- Mtg

 

[haruspex]

What am I solving for? There are many normal forces.

That's exactly the skill you need to develop for such problems - homing in on the useful equations and the important unknowns.

F_net m1x = Fnwl - Fnb sin o
F_net m1y = Fnt - Mbg -Fnb cos o

F_net m2x = Fnb sin o - Fnwr
F_net m2y = Fnb cos o - Mbg

F_net tx = Fnwr - Fnwl
F_net ty = Fnt- Mtg

You don't need the last two equations - they can obviously be deduced from the first four.
For the purposes of finding the torque on the cylinder, which two normal forces are interesting?
If there are any other unknown forces that only occur in one equation, you can safely discard that equation.
That should leave you three equations with three unknowns. Eliminate the uninteresting one and solve for the remaining two.

 

[Samuelriesterer]

I would say eliminate F_m1y. The important forces are Fnw and Fnb. Then:

Fnwl = Fnb sin o = Fnwr
Fnb cos o = Mbg
Fnb = Fnw/sin o
Fnb = Mbg/cos o
Mbg/cos o = Fnw/sin o
Fnw = Mbg * tan o
Mbg = Fnw/tan o
Fnb = (Mbg*tan o)/sin o

I feel like I can go on and on. Am I on the right track?

 

[haruspex]

I would say eliminate F_m1y.

Yes.

The important forces are Fnw and Fnb.

Half right. For figuring out the torque on the cylinder, which two normal forces do you need to know?

Fnwl = Fnb sin o = Fnwr
Fnb cos o = Mbg
Fnb = Fnw/sin o
Fnb = Mbg/cos o
Mbg/cos o = Fnw/sin o
Fnw = Mbg * tan o
Mbg = Fnw/tan o
Fnb = (Mbg*tan o)/sin o

Two of those will do it.

 

[Samuelriesterer]

Ok, how about Fnw and Mbg. And I would choose:

Fnw = Mbg * tan o
Mbg = Fnw/tan o

 

[haruspex]

Ok, how about Fnw and Mbg. And I would choose:

Fnw = Mbg * tan o
Mbg = Fnw/tan o

I was thinking of either of those together with Fnwl = Fnwr.
So, what do you get for torque about the points about which it will potentially tip?

 

[Samuelriesterer]

Torque = F*r = Fnw * d

Wouldn't it be when Fnwl > Fnwr or when Fnw > Mbg?

 

[haruspex]

Torque = F*r = Fnw * d

Wouldn't it be when Fnwl > Fnwr or when Fnw > Mbg?

You correctly identified the pivot point in your diagram. What forces acting on the cylinder have a torque about that point? You must include them all.

 

[Samuelriesterer]

The only forces with x components are Fnw and Fnb, but doesn't Fnb cancel out?

 

[haruspex]

The only forces with x components are Fnw and Fnb, but doesn't Fnb cancel out?

Fnb? Isn't that the normal force between the balls? That's not a force on the cylinder. Did you mean the two normal forces between the balls and the cylinder? Those are equal and opposite as linear forces, but they do not cancel as regards torques.
It's not whether a force has an x component that matters here, it's whether it has a torque on the cylinder about its tipping point. There are three forces that have such a torque, and no two cancel.

 

[Samuelriesterer]

I thought a force directed parallel to the pivot point does not produce any torque so that is why I was looking at the x components of the forces. The only forces on the cylinder are Fnw (normal force on the walls of the cylinder), Fnt (normal force from table), and Mtg (gravitational force on tube)

 

[haruspex]

I thought a force directed parallel to the pivot point does not produce any torque so that is why I was looking at the x components of the forces. The only forces on the cylinder are Fnw (normal force on the walls of the cylinder), Fnt (normal force from table), and Mtg (gravitational force on tube)

Those are the right forces, but you seem to be have a misconception about torques. The torque exerted by a force about a point is the vector cross product of the force and the displacement from the point to the line of action of the force. In scalar terms, that means multiply the magnitude of the force by the perpendicular distance from the point to the line of action. Consequently, the forces which do not exert a torque are those which pass through the point.
Of the four forces you listed, only one passes through the tipping point (when the cylinder is on the point of tipping).

 

[Samuelriesterer]

Ok I understand a little better.

The force that passes through the tipping point exerted on the left wall by the top ball: Fnwl

 

[haruspex]

Ok I understand a little better.

The force that passes through the tipping point exerted on the left wall by the top ball: Fnwl

No. The tipping point is at the base of the left wall. Fnwl does not pass through that point. It is a horizontal force much higher up. This is clear in the diagram you posted right at the start.

 

[Samuelriesterer]

Don't both the normal force by the table Fnt and Mtg both pass through that point?

 

[haruspex]

Don't both the normal force by the table Fnt and Mtg both pass through that point?

At the point of tipping, the normal force from the table will pass through there, but Mtg will pass vertically through the centre of the cylinder (as always). It will miss the pivot point by one cylinder radius.
I worry that you still have a basic misunderstanding somewhere, but I can't pin it down. Are you up to posting another diagram, this time showing just the forces on the cylinder at the point of tipping?

 

[haruspex]

By the way, the given answer is wrong. This can be easily demonstrated by considering what happens as r tends to R/2.

 

[Samuelriesterer]

Yes I believe I can't really grasp Physics in general, let alone Torque, which we kind of glossed over in class. Also, our teacher is not good at all about explaining things and I have to resort to learning from confusing books. Anyway, I have to wrap this assignment up as I have to turn it in tonight. Thanks for your help.