Understanding Engineering Mathematics

mass of a lamina with surface densityρ(x,y)enclosed within the curvey=f(x),the

x-axis and the linesx=a,x=b, are given by

x ̄=

∫b

a

ρxydx

∫b

a

ρydx

y ̄=

1

2

∫b

a

ρy^2 dx

∫b

a

ρydx

If the density is uniform, and such a lamina is rotated about thex-axis, then the centre

of mass of the solid of revolution so formed will lie on thex-axis and thex-coordinate

of the centroid will be given by

x ̄=

∫b

a

xy^2 dx

∫b

a

y^2 dx

(i) Find the positions of the centroids of the following areas (taking

ρ=1):

(a)y=sinx 0 ≤x≤π, y= 0

(b) A semi-circle with radiusa.

(ii) Calculate the positions of the centroids of the solids obtained by

rotating the following curves about the given axes:

(a)x^2 +y^2 =a^2 aboutOx, forx≥ 0 ,y≥ 0.

(b)y^2 = 2 xx= 0 ,x=2 aboutOx.

7.In a rigid body of massMrotating with angular velocityωabout a fixed axis, a particle
Pof massδmat perpendicular distancerfrom the axis would have velocityωrand
kinetic energy^12 δm(ωr)^2. By regarding a continuous body as the limit of a sum of such
particles, its kinetic energy will be

lim

∑ 1

2

ω^2 r^2 δm→

1

2

∫

ω^2 r^2 dm=

1

2

ω^2

∫

r^2 dm

sinceωis the same for all particles of the body. The expression

I=

∫

r^2 dm

is called themoment of inertia(MI) of the body about the axis of rotation. It is usually

given in the formMk^2 wherekis called theradius of gyrationabout the axis. The

MI is also called thesecond momentof the body.

(i) Find the moment of inertia and the radius of gyration of: (a) a uniform

rod lengthaabout perpendicular axis through centre; (b) a disc radius

rabout an axis perpendicular to the disc, through its centre.