Question

Three small spherical masses are located in a plane at the positions shown below. dynamically generated plot The masses are Q=0.300 kg, R=0..600 kg, and S=0.500 kg. Calculate the moment of inertia (of the 3 masses) with respect to an axis perpendicular to the xy plane and passing through x=0 and y=-3. [Since the masses are of small size, you can neglect the contribution due to moments of inertia about their centers of mass.]

i cant figure out how to put the graph up here but the coordinates for the points are as follows

Q=(-5,2)

R=(4,-3)

S=(3,4)

please help!!

Answer 1

by perpendicular axes theorem moment of interia of given axis can be calulated from taking any two mutually perpendicular axis in the xy plane .passing through that axis.

so we will consider line in xy plane passing through (0,-3)

let us take y=-3,

now line perpendicular to this and passing through (0,-3) is x=0.

Itot = I1 + I2

I1 moment of inertia with respect to y=-3

I1= Qr1^2 + Rr2^2 + Sr3^2 ,

for finding perpendicular distanceof point (p,q) from a line ax+by-c=0 use formula mod( ap+bq-c / sqrt(a^2+b^2) )

r1 = perpendicular distance of Q from y=-3 = mod( -2+3 / sqrt(1) ) by substituting -5,2 = 5

similarly

r2=from point (4,-3) that is = mod(-3+3/sqrt(1)) = 0

r3= from point(3,4) that is = mod(4+3/sqrt(1)) = 7

so

I1= 0.3 * 5^2 + 0.6 * 0^2 + 0.5 * 7^2 = 32

now similarly finding I2 with respect to x=0

I2 = Qr4^2 + Rr5^2 + Sr6^2

r4 = mod(-5) = 5 from point (-5,2) to x=0 , r5=4 , r6=3

I2=21.6

Itot = I2+I1 = 32+21.6=53.6