Four particles with masses m 1 = 11 kg, m 2 = 30 kg, m 3 = 8 kg, and m 4 = 39 kg

Question

Four particles with masses m1 = 11 kg, m2 = 30 kg, m3 = 8 kg, and m4 = 39 kg sit on a very light (massless) metal sheet and are arranged as shown below. Find the moment of inertia of this system with the pivot point (a) at the origin and (b) at point P. Assume the rotation axis is parallel to the z direction, perpendicular to the plane of the drawing.

(a) at the origin      kg Four particles with masses m1 = 11 kg, m2 = 30 kg, m3 = 8 kg, and m4 = 39 kg sit on a very light (massless) metal sheet and are arranged as shown below. Find the moment of inertia of this system with the pivot point (a) at the origin and (b) at point P. Assume the rotation axis is parallel to the z direction, perpendicular to the plane of the drawing.

Explanation / Answer

distances of point masses from the origin are:

M1 = (1,1.5)

R1 = 1.8028

M2 = (3,-1.5)

R2 = 3.3541

M3 = (-3,-1)

R3 = 3.1623

M4 = (-0.5,0.5)

R4 = 0.7071

a)

So moment of inertia bout pivot point at origin is M1 * R1^2 + M2 * R2^2 + M3 * R3^2 + M4 * R4^2

= 11 * 3.25 + 30 * 12.25 + 8 * 10 + 39 * 0.5

= 35.75 + 367.5 + 80 + 19.5

= 502.75 kgm^2

b) when pivot point is at p,

we have, square of the distances of masses from the point p as follows

(x-px)^2 + (y-py)^2

(x+2)^2 + (y-2)^2

R1 = 9.25

R2 = 37.25

R3 = 10

R4 = 4.5

So moment of inertia bout pivot point at p is M1 * R1^2 + M2 * R2^2 + M3 * R3^2 + M4 * R4^2

= 11 * 9.25 + 30 * 37.25 + 8 * 10 + 39 * 4.5

= 101.75 + 1117.5 + 80 + 175.5

= 1474.75 kgm^2

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