Find the moment of Inertia?

Question

A compond disk consists of the main disk D and three being disks A,B and C , with their axis parallel to the axis of D and all three being welded to D as shown bellow in the link. The thickness of all four disks is the same, t(a)=t(b)=t(c)=t(d). The radius of the disks A and B are the same , r(a)=r(b)=r(d)/6 . The radius of the disk C is larger , r(c)=r(d)/3. Similarly the mass densities of the disks differ from the main disk. ρ(a)=ρ(b)=2ρ(d). For the disk C , ρ(c)=1.60 ρ(d). Using the parallel-axis theorem, obtain the moment of inertia I of the compound disk about the vertical axis Z which is normal to D and goes through its center. The distance between the centers of disks A,B, and C from the center of the disk D is the same , h(a)=h(b)=h(c).

Data: ρ(d)=3100kg/m^3 , h(a)= 0.810 r(d) , r(d)= 0.620m, t(a)=0.124m.

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Answer 1

The moment of inertia for a disc of radius r and mass m rotating about its center of mass is given by:

I = (mr²)/2

However, if the axis of rotation is displaced by a distance R from the center axis of rotation, the new moment of inertia equals:

I(displaced) = I(center) + mR² = (mr²)/2 + mR²

We are also given the following facts:

ρ(d)=3100kg/m^3
h(a)= 0.810 r(d)
r(d)= 0.620m
t(a)=0.124m

Thickness of all 4 discs are the same: t(a)=t(b)=t(c)=t(d)

The radius of the disks A and B are the same:
r(a)=r(b)=r(d)/6

The radius of the disk C is larger, r(c)=r(d)/3

ρ(a)=ρ(b)=2ρ(d)
ρ(c)=1.60 ρ(d)

The distance between the centers of disks A,B, and C from the center of the disk D is the same , h(a)=h(b)=h(c)

First we calculate the mass of the discs:

M(a) = ρ(a)v(a) = ρ(a)πr²(a)t(a) = 2ρ(d) π(r²(d) / 36)t(a)

= 2(3100)(3.14)(0.620)²(0.124) / 36
= 25.79 kg

M(b) = 25.79 kg
M(c) =82.53 kg
M(d) = 464.21 kg

I(d) = (M(d)r²(d))/2 = (464.21)(0.620)²/2 = 89.22 kg m²

I(a) = (M(a)r²(a))/2 + M(a)h²(a) = M(a) [(r²(d)/72) + (0.810r(d))²]
= 6.64 kg m²

I(b) = 6.64 kg m²

I(c) = (M(c)r²(c))/2 + M(c)h²(c) = M(c) [(r²(d)/18) + (0.810r(d))²]
= (1.762 + 20.814) = 22.58 kg m²

Therefore, I(total) = 89.22 + 2(6.64) + 22.58 = 125.08 kg m²

Answer 2

First, I will compute the mass of each of the disks

mass=ρ*volume
the volume of a disk=pi*r^2*thickness
A=pi*ra^2*ρ(a)*t(a)
B=pi*rb^2*ρ(b)*t(b)
C=pi*rc^2*ρ(c)*t(c)
D=pi*rd^2*ρ(d)*t(d)

A=25.79
B=25.79
C=82.53
D=464.2

The moment of inertia is the sum of the moments

I for D is .25*m*r^2
=.25*464.2*.62^2
=44.61

The other three can be treated as point masses
I for a
ha*ma
=.5022*25.79
=12.95

b is the same
=12.95

c is
=.5022*82.53
=41.45

Sum them all together

=112

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