Calculate the rotational inertia of two 0.250-kg masses located a fixed distance

Question

Calculate the rotational inertia of two 0.250-kg masses located a fixed distance of 0.500 m on opposite sides of an axis of rotation perpendicular to the line joining the masses, (b) Calculate the rotational inertia of a uniform slender rod of length 1.00 m and mass 0.500 kg when it rotates about an axis perpendicular to the rod and through its center, (c) Calculate the moment of inertia of a thin hoop of radius 0.500 m and total mass 0.500 kg if it rotates about an axis perpendicular to the loop and passing through its center, (d) Why are the answers to (a) and (c) the same, but different from (b)?

Explanation / Answer

(a)
The moment of inertia for the system would be each mass's moment of inertia=
I = m * 0.5^2 + m * 0.5^2
I = 2 * 0.25 * 0.5^2 Kg m^2
I =  0.125 Kg m^2

(b)
Moment of Inertia for thin rod = m*l^2/12
I = 0.5 * 1^2 / 12
I = 0.0416 Kg m^2

(c)
I = M*r^2
I = 0.5 * 0.5^2
I = 0.125 Kg m^2

(d)
The moment of inertia of a thin hoop about its central axis is a straightforward extension of the moment of inertia of a point mass since all of the mass is at the same distance R from the central axis, thus value comes out to be same.

While in case of rod , Mass is uniformly distributed along the length of the rod and cannot be replaced with a point mass.

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