Center of mass of cone using cylindrical coordinates

[xdrgnh]

Homework Statement

Set up intergral expression for center of mass of a cone using cylindrical coordinates with a given height H and radius R

Homework Equations

rdrddθdz is part of the inter grand. M/V=D volume of cone is 1/3π(r^2)H

The Attempt at a Solution

dm=Kdv dv=drdθdx K is just a constant because it is uniform density and mass. ∫∫∫zdrdθdz
z=z. Equation for cone in cylindrical coordinates is Z=(H/R)r. The top is bounded by Z=H. The angle is from o to 2π and the radius is square root of H^2+R^2. Once I take this integral I then divide by the same integral but instead there is no extra z in the inter grand. After all is set in done I get the wrong answer. I believe my initial set up might be wrong. Can anyone verify that or give a helpful suggestion thanks.

 

[LCKurtz]

Hard to follow but what did you use for your lower limit on the inside dz integral?

[Edit: Woops I misread your order of integration... wait a sec...Ok, if you integrate dr first, r goes from zero to the cone. What did you use for r on the cone?

 

[xdrgnh]

for r I used the square root of H^2+R^2

 

[LCKurtz]

for r I used the square root of H^2+R^2

But that is a constant. The radius varies as z moves up the cone. Use the equation of the cone to get r in terms of z for your lower limit.

 

[xdrgnh]

I used that for the dr for the dz I used H/R(r) as the lower limit.

 

[LCKurtz]

I used that for the dr for the dz I used H/R(r) as the lower limit.

Your integrals are showing r dr dθ dz in that order. In that order you would integrate z last and its limits must be constant. Order matters so what order are you using?

$$\int_{lower}^{upper}\int_{lower}^{upper}\int_{lower}^{upper} (...)rd?d?d?$$

Quote this in your reply and fill in between the brackets and the question marks with what you did. Then it will be easy to see what you did wrong.

 

[xdrgnh]

Your integrals are showing r dr dθ dz in that order. In that order you would integrate z last and its limits must be constant. Order matters so what order are you using?

$$\int_{(H/R)r}^{H}\int_{0}^{2∏}\int_{0}^{R} (...)rd?d?d?$$

Quote this in your reply and fill in between the brackets and the question marks with what you did. Then it will be easy to see what you did wrong.

$$\int_{0}^{2∏}\int_{0}^{R}\int_{(H/R)r}^{H} (...) zrdzdrdθ$$

 

[LCKurtz]

$$\int_{0}^{2∏}\int_{0}^{R}\int_{(H/R)r}^{H} zrdzdrdθ$$

OK. That integral is correct for your numerator when calculating $\bar z$. When you divide it by the volume you should get $\bar z$ providing you don't make any algebra mistakes. Good luck, I have to go now.