Point particle A has a mass of 295 g and is located at (x, y, z) = (-3.0 cm, 4.5
ID: 1264865 • Letter: P
Question
Point particle A has a mass of 295 g and is located at (x, y, z) = (-3.0 cm, 4.5 cm, 0), point particle B has a mass of 295 g and is at (4.0 cm, 0, 0), and point particle C has a mass of500 g and is at (-4.0 cm, -3.5 cm, 0).
Find the rotational inertia of the system of point particles shown in the figure assuming the system rotates about the following axes.
(a) x-axis
g cm2
(b) y-axis
g cm2
(c) z-axis (The z-axis is perpendicular to the xy-plane and points out of the page.)
g cm2
(d) What are the x and y-coordinates of the center of mass of the system?
xCM = cm
yCM = cm
Additional Materials
Point particle A has a mass of 295 g and is located at (x, y, z) = (-3.0 cm, 4.5 cm, 0), point particle B has a mass of 295 g and is at (4.0 cm, 0, 0), and point particle C has a mass of500 g and is at (-4.0 cm, -3.5 cm, 0). Find the rotational inertia of the system of point particles shown in the figure assuming the system rotates about the following axes. (a) x-axis g cm2 (b) y-axis g cm2 (c) z-axis (The z-axis is perpendicular to the xy-plane and points out of the page.) g cm2 (d) What are the x and y-coordinates of the center of mass of the system? xCM = cm yCM = cm Additional MaterialsExplanation / Answer
a) rotational inertia about x axis
= m1 r1^2 + m2 r2^2 + m3r3^2
= 295*4.5^2 + 295*0 + 500*(-3.5)^2 = 12098.75 g cm^2
b)
rotational inertia about y axis
= m1 r1^2 + m2 r2^2 + m3r3^2
= 295*(-3)^2 + 295*4^2 + 500*(-4)^2 = 15375 g cm^2
c)
rotational inertia about z axis
= rotational inertia about y axis + rotational inertia about x axis
= 15375 + 12098.75 = 27473.75 g cm^2
d) x CM =(m1x1 + ,m2x2 + m3x3 ) /(m1+m2 + m3)
= (295*(-3) + 295*4 + 500*(-4)) / (295 + 295 + 500) = - 1.56 cm
y CM =(m1y1 + m2y2 + m3y3 ) /(m1+m2 + m3)
= (295*(4.5) + 295*0 + 500*(-3.5)) / (295 + 295 + 500) = - 0.39 cm