Metal Base

Metal Base
The mass of a metal cylinder varies jointly as its height and the square of the radius of its base. A cylinder

The mass of a metal cylinder varies together as its height and the square of the radius of its base. A cylinder has a mass of 120 g Find the mass of a second cylinder made of the same metal, three times high, and half the radius of the base of the first.

First we found a good density formula. M is the mass of the cylinder, and V = volume ∴ D (m, V) = the density relative to the mass and volume. D (m, V) = m / V We want to get more detailed then that (do not know the first cylinder volume). So we find the use of a prism volume formula. Let b = base, h = height ∴ V (b, h) = volume in relation to the base and height. V (b, h) = bh. Even more detail (do not know the base). So we use the area of a parabolic sector. Let θ = angle of the sector, r = a shaft, and the other axis ∴ = A (θ, r, a) = area of circle to the line and both radios. b = A (θ, r, a) = θra / 2 We know that the angle of a circle is 2π rad, and that the axes are equal (r = a), so: b (r) = πr ² Plug b (r) V (b, h): V (r, h) = πr ² h Plug V (r, h) D (m, V): D (m, r, h) = m / (πr ² h) This is good to find the density of the first. But we have to find the mass of the second, so clear m. D = m / (πr ² h) / / Multiply by πr ² h and cover. M (D, r, h) = h ² πDr Well, we have our formula. Thus, the first cylinder has a mass of 120 g, but receives no other information, so we use the variables r h yet. And we have to find the mass of the second cylinder, we already have a formula for. We know it is three times greater and has a radius of one half of the original. But we do not know D. But density is constant. Therefore, we will use the density of the first cylinder. D (120 g, r, h) = 120 g / (πr ² h) 120 g / (πr ² h) is the density of the first cylinder. Plug 120 g, 0.5r, and 3h m (d, r, h) m (120 g / (Πr ² h), 0.5r, 3h) is the mass of the second cylinder in relation ship with the first cylinder density, radius and height. m (120 g / (πr ² h) 0.5r, 3h) = (120 g / r ²) ² 3 0.25r Answer: m (120 g / (πr ² h), 0.5r, 3h) = 90 g

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