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A sphere with a given mass floating in water.

  • Thread starter qwerty11
  • Start date
  • #1
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Homework Statement



A spherical aluminum ball of mass 1.26 kg contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate
a.) the outer radius of the ball.
b.) the radius of the cavity.


Homework Equations



I think I have it this time;
d(fluid)gV(object) = m(aluminum)g
d(fluid)(4/3(pi)(r^3)) = m(aluminum)
I solve this for r, and that is the answer for part a.

Next,
V(object) = 4/3(pi)(r^3)
i sub in the radius i found in part a to find V(object)

Next,
V(aluminum) = m(aluminum)/ d(aluminum)
and solve for v aluminum

I then subtract,
V(cavity) = V(object) - V(aluminum)

Finally,
4/3(pi)(r^3) = V(cavity),
and solve for r, this is the answer for b.

The Attempt at a Solution



I think this is the right setup for this problem. My only problem is where do I find the values for aluminum and the fluid?
 

Answers and Replies

  • #2
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Bump.
 
  • #3
Borek
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You mean density values? Tables. For water you can usually use 1 g/mL.

And yes, your approach looks OK.
 
  • #4
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Ok. And I get the mass and desity of aluminum from the periodic table correct?
 
  • #5
Borek
Mentor
28,401
2,800
Depends on what type of information you perdioic table contains, it MAY have density information, it may have not. In general best source for such type of information are engineering handbooks. However, density is listed in billions of places - I suppose it will be on NIST site, in wikipedia and so on.

--
 
  • #6
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Ok, I am really putting some additional wrinkles on the brain because of this problem. Here is what I have now. See what you think.

I think I have it this time;
d(fluid)gV(object) = m(aluminum)g
(1 g/ml)(4/3(pi)(r^3)) = (26.98g)
I solve this for r, and that is the answer for part a.
r=1.86 (what is the unit)
Next,
V(object) = 4/3(pi)(1.86^3)
i sub in the radius i found in part a to find V(object)
V(object)=26.95 (again what units is this)

Next,
V(aluminum) = m(aluminum)/ d(aluminum)
and solve for v aluminum
V(aluminum) = 26.98/2.7gcm^3
V(aluminum) = 9.99 (again what units)

I then subtract,
V(cavity) = V(object) - V(aluminum)
V(cavity) = 26.96 - 9.99
V(cavity) = 16.97 (again what are the units)

Finally,
4/3(pi)(r^3) = V(cavity),
and solve for r, this is the answer for b.
4/3(pi)r^3) = 16.97
r = 1.594 (alas the units are needed here)
 
  • #7
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I believe I was over analyzing it. Check this out.

Mass of sphericle ball (A1)(m)=1.26 kg
Density of Aluminum (Pa)=2.70*10^3 kg/m^3
Density of H20, (Pw) = 1.0*10^3 kg/m^3

(a)
Value of displaceed water = Volume of ball
m/Pw =4/3pi(r)^3
r=(3/4pi*m/pw)^1/3
r=[3/rpi*(1.26kg/1.0*10^3 kg/m^3)]^1/3
r= 6.70cm

B
Since mass of ball = Pa*change in volume
m=Pa*4/3pi(r^3-ri^3)
ri^3=[pa4/3pi(r)^3-m]*3/4pi(pa)
ri^3=[(2700kg/m^3)(4/3pi)(0.067m)^3-(1.26kg)]+3/((4pi)(2700kg/m^3))
ri=5.74 cm


There I think I got it now.
 
  • #8
Borek
Mentor
28,401
2,800
Check what units you use for volume and density, that will tell you units of r. If you mix units (for example density in g/mL and volume in m3) you are either OK if they cancel out, or you are completely off if they don't.

--
 
  • #9
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So my last reply is wrong?
 
  • #10
Borek
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28,401
2,800
No idea - you should check units by yourself. I was taught to do always do it to check if the result makes sense.

From what I see you have used density in kg/m3 so logically answer should be in meters - unless you have converted it to some other length unit.
 
  • #11
ehild
Homework Helper
15,477
1,854
Your result for a.) is wrong. Check the calculation again.

Your formula for the radius of the ball "r=[3/rpi*(1.26kg/1.0*10^3 kg/m^3)]^1/3" is almost OK, but what is "r" in front of pi in the first denominator? And where did the factor 4 go from the previous line?

As for the unit: you divide kg by (kg/m3. What is the result?

ehild
 

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