Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. the distance between any two adjacent masses is ℓ . what is the rotational inertia icm of this system about its center of mass (a point half way between the

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

I = 5mℓ²

are there answer choices to this question? this seems like a question where you have options to choose from.

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.