Four identical particles of mass 0.38 kg each are placed at the vertices of a 3.
ID: 1604712 • Letter: F
Question
Four identical particles of mass 0.38 kg each are placed at the vertices of a 3.0 m 3.0 m square and held there by four massless rods, which form the sides of the square.
(a) What is the rotational inertia of this rigid body about an axis that passes through the midpoints of opposite sides and lies in the plane of the square? 3.41 kg · m2
(b) What is the rotational inertia of this rigid body about an axis that passes through the midpoint of one of the sides and is perpendicular to the plane of the square? 47.88 Incorrect: Your answer is incorrect. kg · m2
(c) What is the rotational inertia of this rigid body about an axis that lies in the plane of the square and passes through two diagonally opposite particles? 3.42 kg · m2
My a and c are correct, I'm just having trouble with part b. I've gotten 18.81 and 47.88 which have both been incorrect. The hint I recieve is: The rotational inertia of a particle is mr2, where m is the mass and r is the perpendicular distance from the given rotation axis.
Explanation / Answer
Total moment of inertia = sum of moments of inertia for each particle about a given axis
(a)
I = 4m(s/2)^2
I = ms^2
I = (0.38 kg)(3.0 m)^2
I = 3.42 kg-m^2
(b)
I = 2m(s/2)^2 + 2m[s^2 + (s/2)^2]
I = 3ms^2 =3*0.38*3^2
I = 10.26 kg-m^2
(c)
I = 2m[s/sqrt(2)]^2
I = ms^2 =0.38*3^2
I = 3.42 kg-m^2
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