Weight - Hollow Steel Ball

[Guest]

A steel ball (hollow) is .5 in in thickness. Outside diameter is 12 in. Determine weight of ball.

Steel weighs .3 lbs. per cu. in.

Volume of sphere - 4/3 x pi x r^3

Area of sphere - 4 x pi x r^2


How do I proceed ? I know the problem is similar to the lead ball problem but the information doesn't give an example for the hollow type.

[Guest]

Please respond - Thank you.

[Guest]

How do I proceed ? I know the problem is similar to the lead ball problem but the information doesn't give an example for the hollow type.

Why do you say this problem is similar to the lead ball problem...????

With the lead ball problem we knew the weight of the large lead ball and found the weight of the small lead balls by using ratio or proportion of their dimensions.

Steel weighs .3 lbs. per cu. in.
Here in this question we know the density of the steel in lbs per cu. in.

So we can work out the weight of the large ball assuming it to be solid.
Then we can work out the weight the inner hollow would be if it were solid
Then if we subtract this from the large ball weight we will get the weight of the shell that is left.

There are other methods we can use as in formulae for volume of hollow sphere etc but the above describes what actually is required....

[Guest]

" Why do you say this problem is similar to the lead ball problem...???? "

The volume formula for a sphere is used.
Trying to determine the weight.

My attempt:

Formula - 4/3 x pi x r^3

4 x 3.1416 x 6 x 6 x 6 = 2714.3424 / 3 = 904.7808 cu. in. ( 6 is the radius of the outside diameter)

4 x 3.1416 x 5.75 x 5.75 x 5.75 = 2388.99045 / 3 = 796.33015 cu. in. (thickness .5 - subtracted from outside diameter 12 = 11.5)

904.7808 - 796.33015 = 108.45065 (difference in large ball from hollow inner)

108.45065 x .3 (lbs. per cu. in.) = 32.535195 lbs.

Where did I make the error(s) ?

[Guest]

4 x 3.1416 x 5.75 x 5.75 x 5.75 = 2388.99045 / 3 = 796.33015 cu. in. (thickness .5 - subtracted from outside diameter 12 = 11.5)

The thickness is all around the sphere......so subtract 0.5 from each end of the outside diameter 12 - 0.5 - 0.5 = 12 - 1 = 11 ins for inside hollow diameter.
BUT your calculations are working using radius, so outer radius is 6 inches and inner radius is 6 - 0.5 = 5.5 inches.

And your formula should be (4 / 3) x Pi x r^3 for the volume of a sphere.

And you are subtracting one formula from the other .....ie..the outer volume - the inner volume.....so it simplifies if a common factor is taken out.

[Guest]

1.333 x 3.1416 x 5.5 x 5.5 x 5.5 = 696.7373 = 697 cu. in.

904.7878 = 905 cu. in.

905 - 697 = 208 (difference)

208 x 0.3 = 62.4 lbs.

[Guest]

62.4 lbs.

Is that the correct answer ?

[Guest]

1.333 x 3.1416 x 5.5 x 5.5 x 5.5 = 696.7373 = 697 cu. in. This is the inner sphere dimensions

You need to work out using the outer sphere dimensions as well

Then subtract them........then use density information to convert to lbs

You need to do these steps below over again.....
=========================
904.7878 = 905 cu. in.
=============================
905 - 697 = 208 (difference)
===============================
208 x 0.3 = 62.4 lbs.
=====================================

[Guest]

1.333 x 3.1416 x 5.5 x 5.5 x 5.5 = 696.7373 = 697 cu. in. (inner)

1.333 x 3 .1416 x 6 x 6 x 6 = 904.5546 = 905 cu. in. (outer)

905 - 697 = 208 (difference)

208 x 0.3(per cu.in.) = 62.4 lbs. weight of ball

[Guest]

Yes

[Guest]

Thanks.

Just for information - What are some other methods to solve you mentioned ?

[Guest]

Take out common factor (4/3xPi) to leave (4/3xPi)(R^3 - r^3) and find shell volume in one calculation then multiply by 0.3 to find weight.
OR (4/3xPix0.3)(R^3 - r^3) to find the weight in one calculation.

This simplifies to (0.4xPi(R^3 - r^3) for this particular question

[Guest]

Thanks again.