A closed 55.0 gallon drum is a 91.0 -long, 59.0 -diameter cylinder with flat ends that is made with 17.0 of sheet metal. If the thickness of the ends is the same as that of the curved sides, what is the moment of inertia about an axis that runs down the center of the drum? The uniform thickness of the sheet metal will allow you to determine the mass of the ends and the mass of the cylinder. What fraction of the surface area of the drum is associated with its ends?
2007-06-14 07:01:12 · 2 answers · asked by physicsdum 1 in Science & Mathematics ➔ Physics
The question doesn't indicate the type of metal
2007-06-14 07:08:18 · update #1
as easy as that sounds what is it made out of? if it was made out of lead or uranium it would weight close to a ton, if it was made out of titanium it would weigh several pounds
2007-06-14 07:06:58 · answer #1 · answered by Flaming Pope 4 · 0⤊ 0⤋
The moment of inertia of a thin-walled hollow cylinder around its central axis is
I = mcyl*r^2
where mcyl is the mass of the thin cylinder and r is its radius.
The moment of inertia of a flat disc around its central axis is
I = 1/2*mdisc*r^2
where mdisc is the mass of the disc and r is its radius.
Once you know the mass of each part of the drum, you simply add up the moments of inertia of each part:
I_total = I_cylindricalshell + 2*I_disc
I_total = mcyl*r^2 + 2*1/2*mdisc*r^2
I_total = mcyl*r^2 + mdisc*r^2
I_total = (mcyl + mdisc)*r^2
This works since the radius of the cylindrical shell is the same as the radius of the disc. Remember that mdisc is the mass of ONE disc.
2007-06-14 07:19:44 · answer #2 · answered by lithiumdeuteride 7 · 0⤊ 0⤋