Question

14. Il The four masses shown in FIGURE EX12.13 are connected by massless, rigid rods. 

a. Find the coordinates of the center of mass. 

b. Find the moment of inertia about a diagonal axis that passes through masses B and D.

Answer 1

a.

Center of mass is given by:

In x-direction

Xcm = (m1x1 + m2x2 + m3x3 + m4x4)/(m1 + m2 + m3 + m4)

In y-direction

Ycm = (m1y1 + m2y2 + m3y3 + m4y4)/(m1 + m2 + m3 + m4)

m1 = 100 g, (x1, y1) = (0,0)

m2 = 200 g, (x2, y2) = (10,0)

m3 = 200 g, (x3, y3) = (0, 8)

m4 = 300 g, (x4, y4) = (10,8)

Now

Xcm = (100*0 + 200*10 + 200*0 + 300*10)/(100+200+200+300)

Xcm = 6.25 cm

Ycm = (100*0 + 200*8 + 200*0 + 300*8)/(100+200+200+300)

Ycm = 5 cm

Center of mass = (6.25 , 5)

b.

Moment of inertia of the system about diagonal BD will be:

I = m1*d1^2 + m2*d2^2 + m3*d3^2 + m4*d4^2

d1 = d3 = distance between mass A or C from BD axis = d (Let)

d2 = d4 = distance between mass B or D from BD axis = 0

here, for 'd',

o city -8 10cm

From triangle ABD,

$\theta$ = arctan(8/10) = 38.66 deg

now, in triangle AOD

AOD = 90 deg

ODA = $\theta$ = 38.66 deg

So, d = 10*sin$\theta$ = 10*sin(38.66 deg)

d = 6.25 cm = 0.0625 m

So,

I = m1*d^2 + m2*0^2 + m3*d^2 + m4*0^2

Using known values:

I = 0.1*0.0625^2 + 0.3*0.0625^2 = 0.00156 kg*m^2

I = 1.56*10^-3 kg*m^2

Let me know if you have any query.