How to calculate the volume of partial cylinder?

JDwg · #1 $Old$ October 23rd, 2009, 11:42 AM

Hello from Finland!
I have a problem how to calculate the volume and the mass centroid of a partial cylinder. Please look at the image below. I have listed the given information there. Variable for a is: a=0...2R

Ive tried to solve it, but I dont know how to do it... All help is appreciated! Thanks!
-Janne

JDwg · #2 $Old$ October 23rd, 2009, 11:48 AM

I found that the solution for a=R is:
V = 2/3*R^2*b
c = R-3/16*PI*R

TKHunny · #3 $Old$ October 23rd, 2009, 03:54 PM

Hyvää on! Miten selvisi?

JDwg · #4 $Old$ October 24th, 2009, 03:56 AM

Quote:

Originally Posted by TKHunny $View Post$

Hyvää on! Miten selvisi?

Solution for a=R I found from a Finnish tablebook called "Rakentajain Kalenteri 2004".
But I need to also find solutions when a is not R...

TKHunny · #5 $Old$ October 25th, 2009, 05:56 AM

I don't understand the drawing.

Is the height 'b' or something greater than 'b'?

Are we using Calculus or only Geometry?

JDwg · #6 $Old$ October 25th, 2009, 08:09 AM

Quote:

Originally Posted by TKHunny $View Post$

I don't understand the drawing.

Is the height 'b' or something greater than 'b'?

Are we using Calculus or only Geometry?

Hi Tk

Height is b. I clarified the drawing below.

Basically Im hoping for an solution for V(olume) and mass (c)entroid in terms of b, a and R.

TKHunny · #7 $Old$ October 26th, 2009, 04:45 PM

Very good. Unfirtunately, I am not at home and have only limited PC access. Perhaps another able source can help you with this before I get around to in on Thursday. Sorry.

TKHunny · #8 $Old$ October 29th, 2009, 12:26 AM

Definitions and Orrientation:

I set it up so that:
1) The y-axis is parallel to the cut.
2) The x-axis is the perpendicular bisector of the cut.
3) The z-axis is parallel to 'b'.
4) The entire structure is in the four quadrants were z > 0.
5) The Origin of the x-y plane is at the center of the circle from the top view.

This has two side effects.
1) The centroid for the y-axis is obviously zero (0), due to symmetry.
2) The equation of the cut-plane is (z-0) = (-b/a)(x-R)
3) The centroid for the x-axis should be < 0. Intuitively clear.

The volume, then, would be:

$2*\int_{-R}^{R-a}\int_{0}^{\sqrt{R^{2}-x^{2}}}\int_{0}^{b}\;dz\;dy\;dx + 2*\int_{R-a}^{R}\int_{0}^{\sqrt{R^{2}-x^{2}}}\int_{0}^{\frac{-b}{a}(x-R)}\;dz\;dy\;dx$

Let's see if you can get that far or if we need a different kind of solution.

TKHunny · #9 $Old$ October 29th, 2009, 07:04 AM

Reality Check:

The Volume is easier than that, ceratinly for a = R, but we'll need this machinery for the Centroid.

For a = R

1) Uncut semi-circular region x < 0

$\frac{1}{2}\cdot \pi R^{2}b$

2) Cut region, x > 0

$\frac{1}{3}\cdot \frac{1}{2}\cdot \pi R^{2} b$

Sure enough, this matches your answer from the book.

TKHunny · #10 $Old$ October 29th, 2009, 09:07 PM

Are you still with me?

In THIS case, since we have symmetry in 'y', we can ignore it and just work with z and x. This makes it rather simple, since the x-z cross section is a simple trapezoid.

$M = \frac{R+(R-a) + 2R}{2} \cdot b$

$M_{x} = \int_{-R}^{R-a}\int_{0}^{b}\;x\;dz\;dx + \int_{R-a}^{R}\int_{0}^{-\frac{b}{R-a}\cdot (x-R)}\;x\;dz\;dx$

Can you do $M_{z}$ ?