Calculating the Radius of a Metallic Ball

In summary, the conversation discussed an object floating on water with a metallic ball attached to it. The object had a length of 6 cm, an area of 12 cm^2, and a density of 0.30g/cm^3. The ball had a density of 19.3g/cm^3. By using the principle of Archimedes, the volume and mass of the object were calculated, and the formula for the volume of a ball with radius r was used to find the radius of the ball, which was 3.1 mm. The conversation highlighted the importance of understanding the principle of Archimedes and having knowledge of equations for calculating volumes of objects in solving physics problems.
  • #1
vinamas
43
1

Homework Statement


A mettalic ball is dangling from a light thread connected to an object that is floating on water the objects length is 6 cm and its area is 12 cm^2 and its density is 0.30g/cm^3 if the floating part of the object is 4 cm what's the radius of the ball (the balls density is 19.3g/cm^3

Homework Equations


Fb=9.81*V*density
V=4/3*r^3*pi
Fg=mg

The Attempt at a Solution


honestly no I idea how (the answer is 3.1 mm)
 
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  • #2
If you really have no idea at all, then you need to review your course notes, a text, and maybe googling the subject you just covered in class.
So what determines how much of an object floats above the water?
 
  • #3
density?
Simon Bridge said:
If you really have no idea at all, then you need to review your course notes, a text, and maybe googling the subject you just covered in class.
So what determines how much of an object floats above the water?
 
  • #4
btw its not like am totally oblivious I have attempted to solve it many times but I don't even get close
 
  • #5
An object will float if the mass of fluid displaced is equal to the mass of the object. This is called "the principle of Archimedes" and you should memorize it.

... in this case you have a composite object.
The mass of an object is it's volume times it's density.
You know an equations for the volumes of various geometric objects.
 
  • #6
Simon Bridge said:
An object will float if the mass of fluid displaced is equal to the mass of the object. This is called "the principle of Archimedes" and you should memorize it.

... in this case you have a composite object.
The mass of an object is it's volume times it's density.
You know an equations for the volumes of various geometric objects.
The object is not completely floating 2 cm of it is submerged in water and that submerged part has the thread that connects the ball to the object
now I can calculate the volume of the object overall floating or the submerged part each with its mass I just don't know how to calculate the mass or volume of the ball
 
  • #7
An object is either floating or it isn't. There is no "completely".
All floating objects are at least partly submerged - it is also possible to float when completely submerged.
The principle of Archimedes works for anything that floats ... everything else sinks.
So you also need to realize that a sunk object, that is not being supported, displaces it's own volume of water. All this should be in your class notes.

You know equations for the volumes of objects ... i.e an object with cross-section area A and height H has volume AH if that object has density p, then it's mass is pAH ... see how I did that?

Do you know the formula for the volume of a ball with radius r?
 
  • #8
Simon Bridge said:
An object is either floating or it isn't. There is no "completely".
All floating objects are at least partly submerged - it is also possible to float when completely submerged.
The principle of Archimedes works for anything that floats ... everything else sinks.
So you also need to realize that a sunk object, that is not being supported, displaces it's own volume of water. All this should be in your class notes.

You know equations for the volumes of objects ... i.e an object with cross-section area A and height H has volume AH if that object has density p, then it's mass is pAH ... see how I did that?

Do you know the formula for the volume of a ball with radius r?
I have solved it thank you I just re arranged some formulas and it worked like magic this is physics I guess
 
  • #9
Well done - what did you do in the end?
 

1. How do you calculate the radius of a metallic ball?

To calculate the radius of a metallic ball, you will need to know the volume and density of the ball. The formula for calculating radius is: radius = (3 * volume) / (4 * pi * density). You can measure the volume of the ball by using a displacement method, and the density can be found in a reference table or by conducting a density experiment.

2. What units should be used when calculating the radius of a metallic ball?

The units used for volume and density should be consistent with each other. For example, if the volume is measured in cubic centimeters (cm^3), then the density should be in grams per cubic centimeter (g/cm^3). The resulting radius will also be in the same unit as the volume, so in this case, it would be in centimeters (cm).

3. Can the radius of a metallic ball be calculated if the mass and diameter are known?

Yes, the radius of a metallic ball can be calculated if the mass and diameter are known. The formula for calculating radius using mass and diameter is: radius = (diameter / 2) * (1 / density) * (mass / pi). Again, the density should be in a consistent unit with the mass and diameter for accurate results.

4. How accurate will the calculated radius of a metallic ball be?

The accuracy of the calculated radius will depend on the accuracy of the measurements used for volume, density, mass, and diameter. The more precise the measurements are, the more accurate the calculated radius will be. It is important to use proper measuring tools and techniques to ensure accurate results.

5. Can the radius of a metallic ball change over time?

The radius of a metallic ball can change over time due to factors like corrosion, wear and tear, or changes in temperature. Therefore, it is important to regularly check and measure the radius of a metallic ball to ensure that it remains within the desired specifications for its intended use.

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