Four identical particles of mass 0.42 kg each are placed at the vertices of a 2.

Question

Four identical particles of mass 0.42 kg each are placed at the vertices of a 2.5 m x 2.5 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? 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? 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? kg m2

Explanation / Answer

part A:

Rotational Inertia = 4 * m (a/2)^2

IR = ma^2

IR = 0.42* 2.5*2.5

IR = 2.625 kgm^2

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part B :

Rotational Inertia IR = 2* m *(a/2)^2 + 2m * (a^2/4 +a^2)

IR = 3 ma^2

IR = 3* 0.42* 2.5^2

IR = 7.875 Kgm^2

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part C:

for Diagonal r = a/sqrt2  

rotational Inertia = IR = 2 m* (a/sqrt2)^2

IR = ma^2

IR = 0.42 *2.5*2.5

IR = 2.625 kgm^2

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