Point particle A has a mass of 280 g and is located at (x, y, z) = (-2.5 cm, 3.5
ID: 1446068 • Letter: P
Question
Point particle A has a mass of 280 g and is located at (x, y, z) = (-2.5 cm, 3.5 cm, 0), point particle B has a mass of 270 g and is at (5.5 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
Explanation / Answer
a) for x- axis,
dA = 3.5cm
dB = 0
dC =3.5cm
Ix = mA dA^2 + mb dB^2 + mC dC^2
Ix = (280 x 3.5^2) + (0 ) + (500 x 3.5^2) = 9555 g cm^2
b) for y- axis,
dA = 2.5cm
dB =5.5cm
dC =4 cm
Ix = mA dA^2 + mb dB^2 + mC dC^2
Ix = (280 x 2.5^2) + (270 x 5.5^2 ) + (500 x 4^2) = 17917.5 g cm^2
c) for z- axis,
dA = sqrt(2.5^2 + 3.5^2) = 4.30 cm
dB = 5.5
dC =sqrt(4^2 + 3.5^2) = 5.32cm
Ix = mA dA^2 + mb dB^2 + mC dC^2
Ix = (280 x 4.30^2) + (270 x 5.5^2 ) + (500 x 5.32^2) = 27495.9 g cm^2
d) x_cm = m1x1 + m2x2 + m3x3 / (m1 +m2 + m3)
= ((280x-2.5) + (270 x 5.5) + (500 x -4)) / (280 + 270 + 500)
= - 1.16 cm
y_cm = = ((280x3.5) + (270 x 0) + (500 x -3.5)) / (280 + 270 + 500)
= - 0.073 cm