Two particles (m1 = 0.20 kg, m2 = 0.30 kg) are positioned at the ends of a2.0-m long rod of negligible mass. what is the moment of inertia ofthis rigid body about an axis perpendicular to the rod and throughthe center of mass? a) 0.48 kg ×m2b) 0.50 kg ×m2c) 1.2 kg ×m2d) 0.80 kg ×m2e) 0.70 kg ×m2

Option (A)

Explanation:

m1 = 0.2 kg

m2 = 0.3 kg

L = 2 m

Let the centre of mass is at a distance d from 0.2 kg.

So, m1 x d = m2 x (L - d)

0.2 x d = 0.3 x (2 - d)

2 d = 6 - 3d

5 d = 6

d = 1.2 m

Moment of inertia about the centre of mass,

I = m1 x d^2 + m2 x (L - d)^2

I = 0.2 x 1.2 x 1.2 + 0.3 x 0.8 x 0.8

I = 0.288 + 0.192

I = 0.48 kg m^2

Thus, option (A) is correct.

high-sugar diet and high cholesterol

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000