Find the rotational inertia of the system of point particles shown in the figure

Question

Find the rotational inertia of the system of point particles shown in the figure assuming the system rotates about the (a) x -axis, (b) y -axis, (c) z -axis. The z -axis is perpendicular to the xy -plane and points out of the page. Point particle A has a mass of 200 g and is located at (x, y, z) = (-5.0 cm, 4.0 cm, 0), point particle B has a mass of 300 g and is at (7.0 cm, 0, 0), and point particle C has a mass of 600 g and is at (-6.0 cm, -6.0 cm, 0). (d) What are the x and y -coordinates of the center of mass of the system? (a) I_x = g middot cm^2 (b) I_y = g middot cm^2 (c) I_z = g middot cm^2 (d) (x, y) = (cm, cm)

Explanation / Answer

here,

m1 = 200 g

m2 = 300 g

m3 = 600 g

a)

the moment of inertia about x axis , Ix = m1 * 4^2 + m2 * 0^2 + m3 * (6^2)

Ix = 200 * 4^2 + 0 + 600 * 6^2 g.cm^2

Ix= 24800 g.cm^2

b)

the moment of inertia about y axis , Iy = m1 * 5^2 + m2 * 7^2 + m3 * (6^2)

Iy = 200 * 5^2 + 300 * 7^2 + 600 * 6^2 g.cm^2

Iy = 41300 g.cm^2

c)

the moment of inertia about z axis , Iz = m1 * (5^2 + 4^2) + m2 * 7^2 + m3 * (6^2 + 6^2)

Iz = 200 * (5^2 + 4^2) + 300 * 7^2 + 600 * (6^2 + 6^2)

Iy = 66100 g.cm^2

d)

the x coordinate of centre of mass , x = (m1 *x1 + m2*x2 + m3*x3 ) /(m1 + m2 + m3)

x = ( 200 * (-5) + 300 * 7 - 600 * 6) /( 200 + 300 +600)

x = - 2.27 cm

the y coordinate of the centre of mass , y = (m1 *y1 + m2*y2 + m3*y3 ) /(m1 + m2 + m3)

y = ( 200 * (4) + 300 * 0 - 600 * 6) /( 200 + 300 +600)

y = - 2.54 cm

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