# Math Help - Find weight of iron ring

1. ## Find weight of iron ring

A cylindrical iron ring has a height of 18 inches. The outside diameter is 120 inches while the inside diameter is 90 inches. What is the weight of the iron ring if 1 cubic feet measures 640 pounds.

Uhmm, this is what I did. I solve for the volume of the two rings(outside and inside). Then I add them together, then convert into cubic feet, then pounds. Is that right?

2. Hello reiward
Originally Posted by reiward

A cylindrical iron ring has a height of 18 inches. The outside diameter is 120 inches while the inside diameter is 90 inches. What is the weight of the iron ring if 1 cubic feet measures 640 pounds.

Uhmm, this is what I did. I solve for the volume of the two rings(outside and inside). Then I add them together, then convert into cubic feet, then pounds. Is that right?
I'm not sure what you mean by the
volume of the two rings(outside and inside)
because you then say you added them. If you mean the total volume (taking the outside dimensions) and then the volume of the space inside (taking the inside dimensions), then you'll need to subtract the second from the first to find the volume of iron.

Then convert, as you did, to cubic feet and hence the weight.

3. Oh I thought I needed to add. Why subtract?

4. Originally Posted by reiward
Oh I thought I needed to add. Why subtract?
Consider the iron cylinder as two cylinders - an outer one which is iron and the inner one which is air

I will denote the outer one as $x_2$ and the inner as $x_1$.

It will also help you considerably if you change your radii $r_1$ and $r_2$ and the height $h$ to feet to avoid unit problems later on.

The volume of a cylinder is given by $V = \pi r^2 h$

If we initially consider the cylinders as solid then we get

$V_2 = \pi r_2 ^2 h$ and $V_1 = \pi r_1 ^2 h$

To find the volume of iron you can take the volume of the air cylinder away from that of the iron cylinder: using the equations above:

$V_{iron} = V_2 - V_1 = \pi r_2^2 h - \pi r_1 ^2 h = \pi h(r_2 ^2 - r_1 ^2)$

All those variables on the right hand side of the last equation are known so sub in the values to find the volume (in cubic feet)

To find the mass you need to use the density given in the question

$\rho = 640 \text{ lb·ft}^{-3}$

The equation linking density and mass is $m = \rho V$

$m_{iron} = 640 \cdot \pi h(r_2 ^2 - r_1 ^2) \text { lb }$

If you or your book have erroneously called the mass the weight then stop here.

To find the weight you must then convert lbs to kg by dividing by 2.2
and then multiply by g. Use whatever value of g the book has (I use 9.81)

$m = 640 \cdot \pi h(r_2 ^2 - r_1 ^2) \text { lb} \times \frac{1}{2.2} \frac{\text { kg }}{\text { lb }} = 640 \cdot \pi h(r_2 ^2 - r_1 ^2) \times \frac{1}{2.2} \text { kg }$

$W = mg = 640g \cdot \pi h(r_2 ^2 - r_1 ^2) \times \frac{1}{2.2} \text { N }$