A brief note on Centre of Mass and the Motion of a System . Read it before a test. Try each question at the end before you open its answer.
The big ideas Centre of mass : x c m = ∑ m i x i ∑ m i x_{cm} = \dfrac{\sum m_ix_i}{\sum m_i} x c m = ∑ m i ∑ m i x i , and the same for y y y , z z z . Nearer the heavier parts; can lie outside the material. Only external forces move it : M a ⃗ c m = F ⃗ ext M\vec{a}_{cm} = \vec{F}_{\text{ext}} M a c m = F ext . Internal forces, however violent, cannot. Momentum of a system : P ⃗ = M v ⃗ c m \vec{P} = M\vec{v}_{cm} P = M v c m ; conserved with no external force. Rigid-body rotation : all particles share ω \omega ω and α \alpha α ; a particle at r r r has v = ω r v = \omega r v = ω r . Rotation equations (steady α \alpha α ): ω = ω 0 + α t \omega = \omega_0 + \alpha t ω = ω 0 + α t , θ = ω 0 t + 1 2 α t 2 \theta = \omega_0t + \tfrac12\alpha t^2 θ = ω 0 t + 2 1 α t 2 , ω 2 = ω 0 2 + 2 α θ \omega^2 = \omega_0^2 + 2\alpha\theta ω 2 = ω 0 2 + 2 α θ . The facts at a glance Two masses a distance d d d apart: x c m = m 2 d m 1 + m 2 x_{cm} = \dfrac{m_2d}{m_1 + m_2} x c m = m 1 + m 2 m 2 d from m 1 m_1 m 1 A hole in a body: treat it as a negative mass Walking L L L along a boat: boat moves m L m + M \dfrac{mL}{m + M} m + M m L Shell bursts in flight: centre of mass keeps the old parabola Rpm to rad/s: multiply by 2 π rad 60 s \dfrac{2\pi\ \text{rad}}{60\ \text{s}} 60 s 2 π rad A worked example A T-shaped sign is made of a 6.0 kg 6.0\ \text{kg} 6.0 kg crossbar, whose centre is 1.2 m 1.2\ \text{m} 1.2 m above the ground, on a 4.0 kg 4.0\ \text{kg} 4.0 kg upright post, whose centre is 0.50 m 0.50\ \text{m} 0.50 m up.
Set up. Both centres lie on one vertical line, so only y y y matters. Average. ∑ m i y i = 6.0 kg × 1.2 m + 4.0 kg × 0.50 m = 9.2 kg m \sum m_iy_i = 6.0\ \text{kg} \times 1.2\ \text{m} + 4.0\ \text{kg} \times 0.50\ \text{m} = 9.2\ \text{kg m} ∑ m i y i = 6.0 kg × 1.2 m + 4.0 kg × 0.50 m = 9.2 kg m , so y c m = 9.2 kg m 10 kg = 0.92 m y_{cm} = \dfrac{9.2\ \text{kg m}}{10\ \text{kg}} = 0.92\ \text{m} y c m = 10 kg 9.2 kg m = 0.92 m . Sense check. It is nearer the heavier crossbar, as expected. Practice with new numbers
A program to try The program finds the centre of mass of the T-shaped sign.
Read it, then run it.
Add a 2.0 kg 2.0\ \text{kg} 2.0 kg lamp at the top, at y = 1.4 m y = 1.4\ \text{m} y = 1.4 m . How far does the centre of mass rise?
Mistakes to avoid Averaging positions without the masses. Each position is weighted by its mass. Dividing by the wrong total. With a hole, divide by the mass that is left. Internal forces moving the centre of mass. They cannot. Turns in the rotation equations. Use radians. Different ω \omega ω for different points. A rigid body has one ω \omega ω ; it is v v v that changes with r r r . Check yourself Masses of 1.0 kg 1.0\ \text{kg} 1.0 kg and 3.0 kg 3.0\ \text{kg} 3.0 kg are 2.0 m 2.0\ \text{m} 2.0 m apart. How far is the centre of mass from the 1.0 kg 1.0\ \text{kg} 1.0 kg mass? Stuck? Show a hint Show the answerA 50 kg 50\ \text{kg} 50 kg person walks 3.0 m 3.0\ \text{m} 3.0 m along a 200 kg 200\ \text{kg} 200 kg raft. How far does the raft move? Stuck? Show a hint Show the answerA wheel turning at 8.0 rad/s 8.0\ \text{rad/s} 8.0 rad/s speeds up steadily to 20 rad/s 20\ \text{rad/s} 20 rad/s in 4.0 s 4.0\ \text{s} 4.0 s . Through what angle does it turn? Stuck? Show a hint Show the answerWhy does a car's speedometer, which counts wheel turns, read high if the tyres are replaced by smaller ones? Stuck? Show a hint Show the answerA rocket in deep space fires its engine. Does the centre of mass of 'rocket plus exhaust gas' move? Stuck? Show a hint Show the answer