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ps3_soln_v2

# ps3_soln_v2 - Physics 105 Problem Set 3 Solutions Problem...

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Physics 105 Problem Set 3 Solutions Problem 1 (3 pts) a) We are asked to find the center of mass of a solid cone of mass M , height L and radius R . Namely, we have to compute the following integral: -→ x cm = cone -→ x dm cone dm = 1 M cone -→ x ρ ( -→ x ) dV In our case the cone has uniform density, so ρ = M/V . Let’s place the base of the cone on the x - y plane and the axis of the cone on the positive z - axis. From rotational symmetry about the cone’s axis, we know that the center of mass must be somewhere on the z - axis, so the 3-D problem can be reduced to a 1-D problem. The radius r of the cone as a function of z is r ( z ) = R - zR/L = R (1 - z/L ). We need to compute the following integral: z cm = 1 M cone zdm ( z ) = 1 M L 0 zρdV ( z ) = 1 M M V L 0 zπr ( z ) 2 dz = πR 2 V L 0 z (1 - z/L ) 2 dz = πR 2 V z 2 2 - 2 z 3 3 L + z 4 4 L 2 L z =0 = πR 2 L 2 12 V . If you don’t happen to know that V ( cone ) = 1 3 base*height, you can calculate it by doing a similar integral as above, just without the extra z in the integrand. V = cone dV ( z ) = L 0 πr ( z ) 2 dz = πR L 0 (1 - z/L ) 2 dz = πR 2 V z - z 2 L + z 3 3 L 2 L z =0 = πR 2 L 3 . 1

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Thus we have z cm = L/ 4, which is independent of the radius R , and the 3-D answer is -→ x cm = (0 , 0 , L/ 4). b) Now let us assume the cone is hollow, with a surface density of σ . Let’s split this problem up into two parts, the upper part which looks like an upside-down ice-cream cone and the base which is disc. We can calculate the center of mass of each and then take the weighted average. The surface area of the upper part can be found by integrating. dl is an infinitesimal segment tangent to the surface of the cone, and dz is its vertical projection. A upper = upper dA ( z ) = upper 2 πr ( z ) dl = 2 πR L 0 (1 - z/L ) L 2 + R 2 L dz = 2 πR L 2 + R 2 L z - z 2 2 L L z =0 = πR L 2 + R 2 . Now we can find its center of mass, which is similar to the integral above but with an extra z in the integrand. z cm,upper = σ M upper upper zdA ( z ) = 1 A upper L 0 z 2 πR (1 - z/L ) L 2 + R 2 L dz = 1 πR L 2 + R 2 2 πR L 2 + R 2 L z 2 2 - z 3 3 L L z =0 = L 3 .
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ps3_soln_v2 - Physics 105 Problem Set 3 Solutions Problem...

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