SME
University Physics I: Mechanics · Problem type
Center of mass of masses on a line
Use this for “A light 1.2 m rod holds 0.4 kg at 0.1 m, 0.6 kg at 0.5 m and 1 kg at 1.1 m. Where is its center of mass?”
Diagram
M = 2 + 3 + 5 = 10 kg.
The balance point is at x = (2 × 0.5 + 3 × 1.5 + 5 × 2.5) ÷ 10 = 1.8 m.
On the fulcrum, the turning effects of the masses on each side cancel.
This is an example. Type your own values to start a new problem.
xcm (m)
=
·
+
·
+
·
M
Formulas
M
=
m₁
+
m₂
+
m₃
10 = 2 + 3 + 5
xcm
=
(
m₁
×
x₁
+
m₂
×
x₂
+
m₃
×
x₃
)
÷
M
1.8 = (2 × 0.5 + 3 × 1.5 + 5 × 2.5) ÷ 10
Assumptions
•
Point masses: each mass sits at one place on the line, and the rod joining them weighs nothing.
•
Places are measured from one origin, + to the right; a place left of the origin is negative.
•
The center of mass is the mass-weighted average place, so Σm(x − xcm) = 0: the rod balances on a pivot there.
Step-by-step
WE KNOW
m₁ = 2 kg, x₁ = 0.5 m, m₂ = 3 kg, x₂ = 1.5 m, m₃ = 5 kg, x₃ = 2.5 m
FIND
total mass (M), center of mass (xcm)
Step 1 · Find total mass (M)
Use M = m₁ + m₂ + m₃
Add the three masses.
M
=
2
+
3
+
5
M
=
10
kg
Step 2 · Find center of mass (xcm)
Use xcm = (m₁ × x₁ + m₂ × x₂ + m₃ × x₃) ÷ M
Weight each place by its mass, add the three, and divide by the total mass.
xcm = (2 × 0.5 + 3 × 1.5 + 5 × 2.5) ÷ 10
xcm
=
1 + 4.5 + 12.5
10
xcm
=
18
10
xcm = 1.8 m
Check
Put the numbers back in. Both sides should match.
10 = 2 + 3 + 5 ✓
1.8 = (2 × 0.5 + 3 × 1.5 + 5 × 2.5) ÷ 10 ✓
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Impulse from a triangular force pulse
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