4.2.3 Centre of Mass — AQA A Level Physics 7408 notes: Define the centre of mass, find it for regular and irregular objects, and use it to analyse stability and toppling.The centre of mass of an object is the point at which its mass can be considered to be concentrated.
- centre of gravity
- x̄ = (m1x1 + m2x2 + ...) / (m1 + m2 + ...)
- stable equilibrium
- tan θ = (b/2) / (h/2) = b/h
- Worked example 1
- Worked example 2
Section 1 of 7Centre of mass and centre of gravity
The centre of mass of an object is the point at which its mass can be considered to be concentrated. The centre of gravity is the point through which the weight of the object appears to act. In a uniform gravitational field, such as near the Earth's surface, the two coincide.
For a symmetrical object of uniform density, the centre of mass is at the geometric centre: the middle of a uniform rod, the centre of a sphere, circle or rectangle.
The weight W = mg can be treated as acting at this single point, which makes moment calculations simple. If you suspend an object from a point, it hangs so that its centre of gravity is directly below the point of suspension, because then the weight has no moment about the pivot.
Section 2 of 7Finding the centre of mass experimentally
For an irregular flat object (a lamina), use the plumb line method:
Using three holes checks the result
- the three lines should meet at a point.
- A uniform rod can be balanced on a knife edge: it balances where the centre of mass is above the pivot.
- Measuring the support forces at the two ends of a body with two balances can also find the centre of mass by moments.
Section 3 of 7Systems of masses
For several point masses lying along a line, the centre of mass is at position
x̄ = (m1x1 + m2x2 + ...) / (m1 + m2 + ...)
where xi are positions measured from a chosen origin.
Section 4 of 7Stability and toppling
An object is in stable equilibrium if, when tilted slightly, it returns to its original position.
- because then the moment of the weight about the pivot edge helps the tilt rather than opposing it.
For a uniform rectangular block of width b and height h resting on a horizontal surface, toppling begins when the block is tilted so that the centre of mass is above the corner, at angle θ where tan θ = (b/2) / (h/2) = b/h.
Section 5 of 7Worked examples
Masses of 2.0 kg, 3.0 kg and 5.0 kg are placed at x = 0, 0.40 m and 1.00 m along a light rod. Find the position of the centre of mass.
x̄ = (2.0 × 0 + 3.0 × 0.40 + 5.0 × 1.00) / (2.0 + 3.0 + 5.0) = (0 + 1.2 + 5.0) / 10.0 = 0.62 m from the first mass.
A bus is 2.4 m wide between wheel contact points, and its centre of mass is 1.6 m above the road. At what angle of tilt does it begin to topple?
tan θ = (w/2)/H = 1.2 / 1.6 = 0.75, so θ = 36.9°.
If passengers sit on the top deck and H rises to 2.0 m, tan θ = 0.60 and the angle falls to 31.0°, making the bus less stable.
A uniform metre rule is balanced at the 50 cm mark. A 100 g mass is hung at the 10 cm mark. At what mark must a 150 g mass be hung to restore balance?
Moments about 50 cm: 100 × (50 − 10) = 150 × d, so d = 4000 / 150 = 26.7 cm.
The 150 g mass goes at the 76.7 cm mark (on the other side of the pivot).
Section 6 of 7Key terms
| Term | Meaning |
|---|---|
| Centre of mass | Point at which the mass of a body can be considered to be concentrated. |
| Centre of gravity | Point through which the weight of a body appears to act. |
| Plumb line | String with a weight that hangs vertically. |
| Stable equilibrium | Returns to its original position when slightly displaced. |
| Lamina | A thin flat sheet. |
| Base | The area of support under an object. |