Physics Olympiad?
Hello dear Gutefragelers,
I have a question regarding a holiday physics assignment. I'm stuck.
The task is as follows:
Two balls of equal size are connected with a long, massless string as shown in the figure and sink at a constant speed.
speed v in water downwards. When the string is cut, the upper ball rises with the same constant speed
speed v back up. The lower sphere is made of aluminum with a density of ρAl = 2700 kg m−3. For the density of water
you can assume the value ρwater = 1000 kg m−3.
Determine the density ρ of the upper sphere.
It would be really great if you could help me
.
Thank you very much for your great support and kind regards!
(3 votes)
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The sink or Ascent speeds are proportional to the density differences to the water.
When sinking, the relevant density is that of the entire arrangement of the two balls, i.e. the mean value 1⁄2(ρ(Al)+ρ)
Of course, only the unknown density ρ plays a role when rising.
Consequently, we get the equation (ρ(Al)+ρ) − ρ(W) = − 2(ρ−ρ(W)) = 2(ρ(W)−ρ)
This can be dissolved according to ρ: ρ=1⁄3( 4ρ(W) − ρ(Al) ) = 433 kg m ̄3
I come to the same result, but I cannot fully understand your bill. How to dissolve ρ(Al)+ρ) − ρ(W) = − 2(ρ−ρ(W)) after ρ in such a way that one can dissolve to ρ=1⁄3(4ρ(W) − ρ(Al) )? I only come to ρ = ρ(W) – 1⁄3ρ(Al)
I think your solution is mbMn. not right:
If the sink or Ascent speeds are the same and the (*) flow resistance FW=1/2*Cw⋅A⋅ρ⋅v^2 or FW=6πηvAct of Stokes, which is left outside, then the following applies:
Fn sink = – Fn raised.
Equations now without factors(g*V).
With 0.433 kg/l,
the “sink force” (g*V*): 2,7+0,433-2=0,677 kg/l and
the “lift force” (g*V*): 0.433-1.0 =0.567 kg/l.
Since these are different, the speeds (without *) would also be different.
I come with my premise to ρ=0.15kg/dm3.
Fn sink = – Fn up, without factor(g*V).
ρAl+ρ-2ρw= -(ρ-ρw) | -ρAl+ρ+2ρw <=>
2ρ=3ρw-ρAl <=>
ρ=(3ρw-ρAl)/2 => ρ=(3*1,00-2,70)/2=0,30/2=0,150 kg/dm3
the “sink force” (g*V*): 2,7+0,150-2=0,850 kg/l and
the “lifting force” (g*V*): 0.433-1.0 =-0.850 kg/l.
How to solve it instead?
What do you mean? I told you my solution.
The law of Stokes must be respected, however, as the balls would become faster and faster when they sink. In the task, however, the density of 433.3 kg/m-3 is correct.
So I have now come to pass with the law of Stroke and the general formula for fr, with the consideration:
fg-fa-fr=0 with v constant, i.e. fr=fg-fa with fr=6*Pi* etha* v, in each case convert to v, and v sink =−v increase.
It is shown to decrease constant for v, increasing that all values are shortened or eliminated except the densities. Finally, the formula for unknown density px=(3p_h2o-p_Al)/2. =>(3-2,7)/2=0,150kg/dm3.
My approach of forces indicated above in the main response was therefore correct, since a constant speed is automatically set and is indirectly given by the resultant increase or drop force which is the same.
Tetesepter007 also claims that the rate of sinking and ascent would be proportional to the density difference to the water.
In his approach, I can’t recognize a Stroke formula.
If I’ve got time and boredom again, I’m trying to figure this out with the resistance forms. I mean preliminary thoughts…
My abdominal feeling tells me that there is nothing much other than my 150 kg/dm^3, because interestingly, in the original task, only constant speed is mentioned, but the speed is not specified… and is itself imagined.
More precisely, if I had any similar buoyancy or sinking power, it would be a ball automatically adjust the V to the same load, since the buoyancy force and sinking force is limited by the same flow resistance which depends on V. But here we have two balls when sinking, which must be said in more detail… In addition “nett”:In the approximation of strut, the flow resistance of the ball is linear to V, but in the original resistance equation square to V…
How to get why would you expect -2pw to fall and -pw to rise? In principle, I do not think the approach is wrong, but I do not think why these figures are calculated.
You’ve got two bullets at the sink, that’s, 2 displaced volumes, but only one at the rise.