Evari5te
- 10
- 4
- Homework Statement
- K.A Stroud Engineering Mathematics 8th edition Kindle version Chapter 2.21 Further problems #1
A plane figure is enclosed by the x-axis and the curve ##y=a\sin x## between ##x=0## and ##x=\pi##
Show that the radius of gyration ##k##, of the figure about the x-axis is given by ##\frac{a.\sqrt[]{2}}{3}##
- Relevant Equations
- Radius of gyration, ##k = \sqrt[]{\left(\frac{I}{A}\right)}## where
##I## is the Area Moment of Inertia given by integrating the square of the distance ##y## from the x-axis to each infinitesimal element of area, ##dA## of the figure, multiplied by the area element itself.
##I=\int_0^\pi y^2dA##
##\therefore I = \int_0^\pi a^2 \sin^2 x dA##
##dA = y.dx = a \sin x dx##
##\therefore I = \int_0^\pi a^3 \sin^3 x dx = a^3 \int_0^\pi \sin^2 x.\sin x dx##
The closest I can get is ##k=\frac{\sqrt{2}a}{\sqrt{3}}## as follows:
##I=\int_0^\pi y^2dA##
##\therefore I = \int_0^\pi a^2 \sin^2 x dA##
##dA = y.dx = a \sin x dx##
##\therefore I = \int_0^\pi a^3 \sin^3 x dx = a^3 \int_0^\pi \sin^2 x.\sin x dx##
Let
##u=\sin^2x##, ##\therefore \frac{du}{dx} = 2\sin x \cos x##
##\frac{dv}{dx}=\sin x##
##\therefore v = \int \sin x dx =-\cos x +C##
Integrating by parts
##I=a^3 \left\{uv - \int_0^\pi v\frac{du}{dx} dx \right\} ##
##I = a^3 \left\{ \left[\sin^2 x \left(- \cos x \right) \right]_0^\pi - \int_0^\pi \left(- \cos x \right).2\cos x.\sin x dx \right\} ##
Because ##\sin 0=0## and ##\sin \pi =0##, ##\left[\sin^2 x \left(- \cos x \right) \right]_0^\pi = 0 - 0##
leaving
##I = a^3 \left\{ - \int_0^\pi \left(- \cos x \right).2\cos x.\sin x dx \right\} ##
Taking the ##-1## multiplier from the ##\left(- \cos x \right)## term and ##2## multipler from the ##2\cos x.\sin x## term outside of the integral (so negating the minus) gives
##I=+2a^3 \int_0^\pi \left(\cos x \right).\cos x.\sin x dx = +2a^3 \int_0^\pi \cos^2 x.\sin x dx##
Using the identity ##\cos^2 x = 1 - \sin^2 x##
##I =2a^3 \int_0^\pi \left(1-\sin^2 x \right).\sin x dx ##
##I=2a^3 \left\{ \int_0^\pi \sin x dx - \int_0^\pi \sin^3 x dx \right\}##
##I= 2a^3 \left[-\cos x \right]_0^\pi - 2a^3 \int_0^\pi \sin^3 x dx ##
The term ##a^3 \int_0^\pi \sin^3 x dx## on the right hand side is now the same as the original integral so
##I = 2a^3\left[-\cos x \right]_0^\pi - 2I##
##\therefore 3I = 2a^3\left[-\cos x \right]_0^\pi##
##\cos \pi = -1 \therefore -\cos \pi = 1 ##
##\cos 0 = 1 \therefore -\cos 0 = -1##
##\therefore 3I = 2a^3\left[1 - \left(-1\right) \right] = 2a^3\left(2\right) = 4a^3##
##\therefore I = \frac{4a^3}{3}##
##A## is the area of the figure,
##A=\int_0^\pi a \sin x dx = a\left[-cos x\right]_0^\pi = 2a##
## k=\sqrt[]{\frac{I}{A}}=\sqrt{\frac{4a^3}{3.(2a)}} = \sqrt{\frac{2a^2}{3}}=\frac{\sqrt{2}a}{\sqrt{3}}##
Am I wrong or is there a typo in the book?
##I=\int_0^\pi y^2dA##
##\therefore I = \int_0^\pi a^2 \sin^2 x dA##
##dA = y.dx = a \sin x dx##
##\therefore I = \int_0^\pi a^3 \sin^3 x dx = a^3 \int_0^\pi \sin^2 x.\sin x dx##
Let
##u=\sin^2x##, ##\therefore \frac{du}{dx} = 2\sin x \cos x##
##\frac{dv}{dx}=\sin x##
##\therefore v = \int \sin x dx =-\cos x +C##
Integrating by parts
##I=a^3 \left\{uv - \int_0^\pi v\frac{du}{dx} dx \right\} ##
##I = a^3 \left\{ \left[\sin^2 x \left(- \cos x \right) \right]_0^\pi - \int_0^\pi \left(- \cos x \right).2\cos x.\sin x dx \right\} ##
Because ##\sin 0=0## and ##\sin \pi =0##, ##\left[\sin^2 x \left(- \cos x \right) \right]_0^\pi = 0 - 0##
leaving
##I = a^3 \left\{ - \int_0^\pi \left(- \cos x \right).2\cos x.\sin x dx \right\} ##
Taking the ##-1## multiplier from the ##\left(- \cos x \right)## term and ##2## multipler from the ##2\cos x.\sin x## term outside of the integral (so negating the minus) gives
##I=+2a^3 \int_0^\pi \left(\cos x \right).\cos x.\sin x dx = +2a^3 \int_0^\pi \cos^2 x.\sin x dx##
Using the identity ##\cos^2 x = 1 - \sin^2 x##
##I =2a^3 \int_0^\pi \left(1-\sin^2 x \right).\sin x dx ##
##I=2a^3 \left\{ \int_0^\pi \sin x dx - \int_0^\pi \sin^3 x dx \right\}##
##I= 2a^3 \left[-\cos x \right]_0^\pi - 2a^3 \int_0^\pi \sin^3 x dx ##
The term ##a^3 \int_0^\pi \sin^3 x dx## on the right hand side is now the same as the original integral so
##I = 2a^3\left[-\cos x \right]_0^\pi - 2I##
##\therefore 3I = 2a^3\left[-\cos x \right]_0^\pi##
##\cos \pi = -1 \therefore -\cos \pi = 1 ##
##\cos 0 = 1 \therefore -\cos 0 = -1##
##\therefore 3I = 2a^3\left[1 - \left(-1\right) \right] = 2a^3\left(2\right) = 4a^3##
##\therefore I = \frac{4a^3}{3}##
##A## is the area of the figure,
##A=\int_0^\pi a \sin x dx = a\left[-cos x\right]_0^\pi = 2a##
## k=\sqrt[]{\frac{I}{A}}=\sqrt{\frac{4a^3}{3.(2a)}} = \sqrt{\frac{2a^2}{3}}=\frac{\sqrt{2}a}{\sqrt{3}}##
Am I wrong or is there a typo in the book?