Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 40P from Chapter 9 from Hibbeler's Engineering Mechanics.
Problem 40P
Step-by-Step Solution
We are given the curve equation of paraboloid as ${z^2} = 4y$.
We are asked to locate the centroid $\bar y$ of the paraboloid.
Step 2
We can consider the differential elemental shape of the disk of thickness $dy$ which is located at a distance of y from z-axis.
Here, z is the radius of a disk.
The curve equation is given as:
\[{z^2} = 4y\]The volume of the element can be given as:
\[dV = \left( {\pi {z^2}} \right)dy\] … (1)To calculate the centroid $\bar y$ of the paraboloid, we have:
\[\bar y = \frac{{\int\limits_0^4 {ydV} }}{{\int\limits_V {dV} }}\] …. (2)Step 3
To calculate the integral of the elemental volume, we need to integrate equation (1) with the limit $0$ to $4$ as:
\begin{array}{c} \int\limits_V {dV} = \int\limits_0^4 {\left( {\pi {z^2}} \right)} dy\\ \int\limits_V {dV} = \int\limits_0^4 {\left( {4\pi y} \right)} dy\\ \int\limits_V {dV} = 4\pi \left( {\frac{{{y^2}}}{2}} \right)_0^4 \end{array}Substitute the limits in y, we get:
\begin{array}{c} \int\limits_V {dV} = 4\pi \left[ {\frac{{{{\left( 4 \right)}^2}}}{2} - 0} \right]\\ \int\limits_V {dV} = 32\pi \end{array}Substitute the values in equation (2), we get:
\[\begin{array}{c} \bar y = \frac{{\int\limits_0^4 {y\left( {\pi {z^2}} \right)dy} }}{{\left( {32\pi } \right)}}\\ \bar y = \frac{{\int\limits_0^4 {y\left( {4\pi y} \right)dy} }}{{\left( {32\pi } \right)}}\\ \bar y = \left( {\frac{1}{8}} \right) \times \left( {\frac{{{y^3}}}{3}} \right)_0^4 \end{array}\]Substitute the limits in y, we get:
\[\begin{array}{c} \bar y = \left( {\frac{1}{8}} \right) \times \left[ {\frac{{{{\left( 4 \right)}^3}}}{3} - 0} \right]\\ \bar y = \left( {\frac{1}{8}} \right) \times \left( {21.33} \right)\\ \bar y = 2.67\;{\rm{m}} \end{array}\]