trainee engineer
New member
- Joined
- May 16, 2022
- Messages
- 11
Good Afternoon,
I have been working on a problem that requires the determination of the centre of mass of a truncated parabola rotated around the Y axis.
The question states:
The curve [math]y=x^2-1[/math] is to be rotated about the y-axis, for 1 ≤ x ≤ 3 . This forms a bowl shaped body, of a constant density.
a) Calculate the Volume of the bowl
b) Find the co-ordinates of the centre of mass.
Using the rule for volume of a rotation, [math]V= \pi \int x^2 dy[/math] as it rotates around the y axis, I was able to ascertain the volume, [math] V = 40 \pi [/math]
I am unsure of my result for the centre of mass
My rule is [math]\overline{y}=\frac{1}{V}\int y \cdot A(y) dy[/math]
As I am taking circular cross sections, parallel to the x axis, the Area of the circles is [math]A=\pi x^2[/math], as the radius is always
equal to x.
rearranging the original curve gives [math]x=\sqrt{y+1}[/math], so I have substituted that into my formula.
when x = 1, y = 0 and when x = 3, y = 8, I have used these as my upper and lower limits for integration.
The working I have is:
[math]\overline{y}=\frac {1} {V} \int_{0}^{8} y \cdot A(y) dy[/math][math]\overline{y}=\frac {1} {40 \pi } \int_{0}^{8} y \pi (\sqrt{y+1} )^2 dy[/math][math]\overline{y}=\frac {\pi } {40 \pi } \int_{0}^{8} y (y+1) dy[/math][math]\overline{y}=\frac {1} {40} \int_{0}^{8} y^2+ y dy[/math][math]\overline{y}=\frac {1} {40} \left[\frac{y^3}{3}+\frac{y^2}{2}\right]_0^8[/math]
Subbing in gives me
[math]\overline{y}=\frac{52}{15}[/math]
so the co-ordinates of the centre of mass will be [math](0,\frac{52}{15},0)[/math]
It does not seem right that the centre of mass is less than half of the height, given that the bowl is concave up.
Any thoughts?
I have been working on a problem that requires the determination of the centre of mass of a truncated parabola rotated around the Y axis.
The question states:
The curve [math]y=x^2-1[/math] is to be rotated about the y-axis, for 1 ≤ x ≤ 3 . This forms a bowl shaped body, of a constant density.
a) Calculate the Volume of the bowl
b) Find the co-ordinates of the centre of mass.
Using the rule for volume of a rotation, [math]V= \pi \int x^2 dy[/math] as it rotates around the y axis, I was able to ascertain the volume, [math] V = 40 \pi [/math]
I am unsure of my result for the centre of mass
My rule is [math]\overline{y}=\frac{1}{V}\int y \cdot A(y) dy[/math]
As I am taking circular cross sections, parallel to the x axis, the Area of the circles is [math]A=\pi x^2[/math], as the radius is always
equal to x.
rearranging the original curve gives [math]x=\sqrt{y+1}[/math], so I have substituted that into my formula.
when x = 1, y = 0 and when x = 3, y = 8, I have used these as my upper and lower limits for integration.
The working I have is:
[math]\overline{y}=\frac {1} {V} \int_{0}^{8} y \cdot A(y) dy[/math][math]\overline{y}=\frac {1} {40 \pi } \int_{0}^{8} y \pi (\sqrt{y+1} )^2 dy[/math][math]\overline{y}=\frac {\pi } {40 \pi } \int_{0}^{8} y (y+1) dy[/math][math]\overline{y}=\frac {1} {40} \int_{0}^{8} y^2+ y dy[/math][math]\overline{y}=\frac {1} {40} \left[\frac{y^3}{3}+\frac{y^2}{2}\right]_0^8[/math]
Subbing in gives me
[math]\overline{y}=\frac{52}{15}[/math]
so the co-ordinates of the centre of mass will be [math](0,\frac{52}{15},0)[/math]
It does not seem right that the centre of mass is less than half of the height, given that the bowl is concave up.
Any thoughts?
