Optimizing Mass for Vertical Rod Buoyancy in Unknown Fluid

[timetraveller123]

Homework Statement

A 0.75-m rod has a uniform linear mass density of λ. A small mass m with negligible volume is attached to one end of the rod. The rod with the attached mass is placed in a container of unknown fluid and after oscillating briefly, comes to rest at its equilibrium position. At equilibrium, the rod floats vertically with 2/3 of its length submerged and mass m in the fluid. If the rod were fully submerged it would displace 7.5×10^-4 kg of fluid.

(a) What is the maximum value that the mass m can have?
(b) What is the minimum value that the mass m can have?
(c) Sketch a graph that shows the values of λas a function of m

Homework Equations

f_buoyancy = g ρ v

The Attempt at a Solution

f_boyant = f_weight

g 5 x 10^-4 = g (m + 0.75 λ)

how is m supposed to have maximum and minimum values for this equation

 

[BvU]

Think of the rest position of the rod. Is there a degree of freedom you overlooked ? Make a drawing.

 

[timetraveller123]

isn't there only one degree of freedom as the question specifies it is vertically aligned and and the mass is at the bottom

 

[BvU]

My mistake. I thought there was an angle left over, but you are right. In that case I don't see why m isn't fully determined and we're in the same boat

Maybe we should start calculating m and see where we end up ...

 

[timetraveller123]

but we need to know λ for that

 

[BvU]

Gather the equations - use the symbol $\lambda$

 

[BvU]

Oops, I see it's there already. 5 x 10^-4 = m + 0.75 λ

Well then all I can think of is a liberal interpretation of 'floats vertically': that it means 'in a vertical plane, but at an angle with the liquid surface'

 

[timetraveller123]

so how is that supposed to help even if it floats at an angle how are we supposed to calculate the fluid it displaces

 

[BvU]

I think that stays at 2/3 of the volume of the rod. I was thinking about the of the given 'vertical orientation': if it wants to be stable, there is a condition. Perhaps that sets bounds on m ?

 

[haruspex]

I think that stays at 2/3 of the volume of the rod. I was thinking about the of the given 'vertical orientation': if it wants to be stable, there is a condition. Perhaps that sets bounds on m ?

Quite so, it is a question about stability.
@vishnu 73 , suppose the rod is not quite vertical, but at some small angle to the vertical. What relationship between the forces will lead it to rotate towards vertical?

Edit: to clarify, the stability consideration is for finding the minimum value of m. The maximum value is simpler.

 

[timetraveller123]

Quite so, it is a question about stability.
@vishnu 73 , suppose the rod is not quite vertical, but at some small angle to the vertical. What relationship between the forces will lead it to rotate towards vertical?

Edit: to clarify, the stability consideration is for finding the minimum value of m. The maximum value is simpler.

i did what you said but i got the opposite results meaning i got maximum value for when the rod is tilted and minimum value for m when the rod is perfectly vertical my calculation are as follows

when the rod is perfectly vertical:

f_b = f_weight
5 x 10^-4 g = mg + λ 0.75g
5 x 10^-4 = m + λ 0.75
5 x 10^-4 - λ 0.75 = m this is the minimum value of m

when the rod is tilted i used torque about the point where the rod intersects the water surface = 0
hence,

f_b 1/3 L= f_weight 1/6 L

2x(5 x 10^-4 g) = (mg + λ 0.75g)
10^-3 = m + 0.75λ
10^-3 - 0.75λ = m this is the maximum value for m i got

did i do something wrong please help!

 

[haruspex]

when the rod is perfectly vertical:
5 x 10-4 = m + λ 0.75

That is the condition for it to float 2/3 submerged, vertical or not.

when the rod is tilted i used torque about the point where the rod intersects the water surface = 0
hence,
f_b 1/3 L= f_weight 1/6 L

The mass centre of the rod is 1/6 below the surface, but not the mass m.

 

[timetraveller123]

ah yes forgot about that
so working it out again i got
for tiled rod:

f_b = f_weight
5 x 10^-4(1/3) = (2/3)m + 0.75(1/6)λ
10^-3 -0.75λ = 4m
m = (10^-3 -0.75λ)/4

is this correct is this lesser than the m for the just floating one
and why is it that the tiled rod can have a lower value for m than the just floating rod.

 

[haruspex]

ah yes forgot about that
so working it out again i got
for tiled rod:

f_b = f_weight
5 x 10^-4(1/3) = (2/3)m + 0.75(1/6)λ
10^-3 -0.75λ = 4m
m = (10^-3 -0.75λ)/4

is this correct is this lesser than the m for the just floating one
and why is it that the tiled rod can have a lower value for m than the just floating rod.

Your equations are right now, but you are not interpreting them correctly.

5 x 10^-4 - λ 0.75 = m this is the minimum value of m

No, it's not the minimum value of m. It is a relationship between m and λ that must be true, given that it floats 2/3 submerged.

m = (10^-3 -0.75λ)/4

This is the relationship between m and λ when it is borderline whether it will be vertical or not (and if not, it will float horizontally; tilted is unstable).
But you do not care about λ, so combine the two equations to eliminate it.

Similarly, for the maximum value of m, the question wants the absolute maximum over all possible values of λ. This is very easy.

 

[timetraveller123]

huh i see what you are saying but when i combine the equations
m = (10-3 -0.75λ)/4
5 x 10-4 - λ 0.75 = m
and solving simultaneously i get one solution and not a range for me to find the minimum
i get m = 10^-3/(4+5x10^-4) this is just 2/8001

and for the maximum value of value the only thing i can think of is λ = 0;
but how that effectively is no rod in which case the m = 5x10^-4

is my above working correct ?
i think i have a problem understanding this question the question already specifies the rod is vertical then i don't understand the need for calculating m when it is horizontal please help me explain what the question is asking thanks!

 

[haruspex]

i get one solution and not a range for me to find the minimum

The equations give you the minimum.

i get m = 10-3/(4+5x10-4)

That's not what I get. Please check your working. If you still get the same, please post all your steps.

i don't understand the need for calculating m when it is horizontal

You did not calculate m for being horizontal. Your second equation is for the rod being on the borderline between remaining vertical and falling over. If it does fall over it will go horizontal. There is no stable solution between the two.

 

[timetraveller123]

sorry my bad about the equation i was careless

i now get m = (10^-3 - 5x10^-4)/3 = 1/6000

why can't the rod be tilted why must it be vertical or horizontal
i am sorry i still don't get it but why does the above equation get me the minimum value of m
and so is my maximum value in the previous post correct
and if its not too much too ask please walk me through without equation what the is logic behind this entire question that would be very helpful

 

[haruspex]

= (10-3 - 5x10-4)/3 = 1/6000

Yes.

why must it be vertical or horizonta

Your second equation says the torque is in perfect balance, that it will be stable at any angle. If m is a fraction more, the net torque will push it towards the vertical, and it will keep going until it gets there. If m is a fraction less, the net torque will push it away from the vertical, and it will keep going until it is horizontal. Note that your equation did not depend on the angle.

why does the above equation get me the minimum value of m

Because if m were any less the vertical arrangement would be unstable.

is my maximum value in the previous post correct

No. It needs to be expressed independently of λ. At what value of m is it impossible for the rod to float 2/3 immersed no matter how small λ is?

 

[timetraveller123]

ok now i get the minimum value part but idont understand the maximum value part i don't see how it is supposed to be simpler the best i can do is

m = 5x10^-4 - 0.75λ

i don't see you can get rid of the λ

 

[haruspex]

ok now i get the minimum value part but idont understand the maximum value part i don't see how it is supposed to be simpler the best i can do is

m = 5x10^-4 - 0.75λ

i don't see you can get rid of the λ

What is the smallest possible λ?

 

[timetraveller123]

is it zero

 

[timetraveller123]

or wait should i solve the above two equations for λ instead of m

 

[haruspex]

or wait should i solve the above two equations for λ instead of m

No, it asks for the min and max of m. What is max m if min λ is zero?

 

[timetraveller123]

5x10^-4?

 

[haruspex]

5x10^-4?

Yes.

 

[timetraveller123]

but i said that on post #15 lol anyways thanks ! for the help

 

[haruspex]

but i said that on post #15 lol anyways thanks ! for the help

So you did. Sorry, I either missed that or later forgot.

 

[timetraveller123]

just asking in your own opinion is the question badly phrased ?

 

[haruspex]

just asking in your own opinion is the question badly phrased ?

It could have been clearer that the min and max values of m were to be over all possible values of λ.

 

[timetraveller123]

ok thanks