How fast does the centre of mass move?
The first lesson made a promise: the centre of mass moves like a single particle carrying the whole mass. Four lessons have located the point. This lesson proves the promise, and it needs nothing more than the definition and two derivatives. Keep one condition in mind throughout: the system keeps the same particles, so its total mass \(M\) does not change.
\[ M\vec r_{cm} = m_1\vec r_1 + m_2\vec r_2 + \cdots + m_n\vec r_n \]
Why. the definition of the centre of mass, multiplied by the total mass \(M\).
\[ M\dfrac{d\vec r_{cm}}{dt} = m_1\dfrac{d\vec r_1}{dt} + m_2\dfrac{d\vec r_2}{dt} + \cdots \]
Why. differentiate both sides with respect to time; every mass is constant.
\[ M\vec v_{cm} = m_1\vec v_1 + m_2\vec v_2 + \cdots + m_n\vec v_n \]
Why. the rate of change of each position vector is the velocity of that particle.
\[ M\vec v_{cm} = \vec p_1 + \vec p_2 + \cdots + \vec p_n = \vec P \]
Read it. each \(m_i\vec v_i\) is the momentum \(\vec p_i\) of a particle, and their vector sum is the total momentum \(\vec P\).
The velocity of the centre of mass is therefore the mass-weighted average of the velocities. The last line says more. The total momentum of a system, however its parts move, equals the total mass times the velocity of one point, the centre of mass. A spinning bat, a swarm of bees and a family in a car each have a total momentum \(M\vec v_{cm}\).
A \(2\,\text{kg}\) puck moves with velocity \((3\hat i + 4\hat j)\,\text{m/s}\) on a smooth table, and a \(3\,\text{kg}\) puck moves with velocity \((-2\hat i + \hat j)\,\text{m/s}\). Find the velocity of their centre of mass and their total momentum.
Work one component at a time, with \(M = 5\,\text{kg}\).
Along x: \(v_{cm,x} = (2 \times 3 + 3 \times (-2))/5 = (6 - 6)/5 = 0\).
Along y: \(v_{cm,y} = (2 \times 4 + 3 \times 1)/5 = 11/5 = 2.2\,\text{m/s}\).
So \(\vec v_{cm} = 2.2\hat j\,\text{m/s}\), and \(\vec P = M\vec v_{cm} = 11\hat j\,\text{kg m/s}\). The x momenta of the two pucks cancel, although both pucks move along x.
A \(4\,\text{kg}\) trolley moves right at \(3\,\text{m/s}\) and a \(6\,\text{kg}\) trolley moves left at \(2\,\text{m/s}\) on the same track. What is the velocity of their centre of mass?
Only outside forces count
Differentiate once more. The rate of change of each velocity is an acceleration, so \(M\vec a_{cm} = m_1\vec a_1 + m_2\vec a_2 + \cdots\). By Newton's second law, each \(m_i\vec a_i\) is the net force on particle \(i\). So \(M\vec a_{cm}\) is the sum of all the forces on all the particles of the system.
Sort these forces into two kinds. An internal force is exerted by one particle of the system on another. Examples are the push of a spring between two blocks and the tension in a rope joining them. The pressure of the gas in an exploding shell is another. An external force comes from outside the system: the Earth's gravity, the push of a hand, the friction of a floor.
\[ M\vec a_{cm} = \sum_i m_i\vec a_i = \sum_i \vec F_i \]
Why. differentiate \(M\vec v_{cm} = \sum_i m_i\vec v_i\), then use Newton's second law for each particle.
\[ \sum_i \vec F_i = \sum \vec F_{\text{int}} + \sum \vec F_{\text{ext}} \]
Why. every force on every particle is either internal or external.
\[ \sum \vec F_{\text{int}} = \vec 0 \]
Why. Newton's third law: if particle 1 pushes particle 2 with \(\vec F_{12}\), then 2 pushes 1 with \(\vec F_{21} = -\vec F_{12}\). Both are internal, so they cancel in the sum.
\[ M\vec a_{cm} = \vec F_{ext} \]
Read it. only the net external force is left.
The centre of mass of a system moves as a particle of the total mass \(M\) under the net external force: \(M\vec a_{cm} = \vec F_{ext}\). Internal forces, however large, cannot change the motion of the centre of mass.
This is the answer to the question left open in the first lesson. When we wrote \(F = ma\) for a block, the block was a system of countless particles, and \(a\) was the acceleration of its centre of mass. The earlier chapters were right to treat the block as one particle: its centre of mass is that particle.
The theorem is about the centre of mass only. Internal forces cannot move the centre of mass, but they certainly change how each part moves. A spring between two blocks makes each block speed up and slow down. It does nothing to their centre of mass.
Two blocks and a spring, as one body
Two \(1\,\text{kg}\) blocks rest on a smooth floor, joined by a light spring. A steady force \(F = 2\,\text{N}\) pulls the front block. The front block starts at once, the spring stretches, and the stretched spring then drags the rear block. The spring force keeps changing, so each block surges and slows. Solving for each block needs the spring constant and some hard mathematics.
The centre of mass needs none of that. The spring force is internal, so it cancels. The floor is smooth, and the weights balance the normal forces, so the only unbalanced external force is \(F\). Hence \(a_{cm} = F/M = 2/2 = 1\,\text{m/s}^2\), constant from the first instant. In Fig 5.1 the blue dots spread out evenly, like the dots of one body under steady acceleration. The orange dots of block B bunch and stretch as the spring tugs back.
A \(3\,\text{kg}\) block and a \(1\,\text{kg}\) block, joined by a light spring, rest on a smooth floor. A steady horizontal force of \(12\,\text{N}\) pulls the \(1\,\text{kg}\) block away from the other. While the spring stretches and the blocks jerk, what is the acceleration of their centre of mass, in m/s²?
The Atwood machine, as one body
The chapter on Accelerated Objects solved the Atwood machine block by block. Now view it as one system: the two blocks and the string. Take blocks of mass \(m\) and \(3m\) over a light pulley clamped to the ceiling. The heavy block goes down and the light one rises, each with acceleration \(a = (3m - m)g/4m = g/2\).
In the Atwood machine of Fig 5.2, find the acceleration of the centre of mass of the two blocks. Also find the force with which the clamp holds the pulley. The pulley and the string are light.
Take the system as the two blocks, the string and the pulley, with downward positive. The external forces are the two weights, \(mg + 3mg = 4mg\) down, and the clamp force \(N\) up.
\(a_{cm} = \dfrac{3m \times (g/2) + m \times (-g/2)}{4m}\), and \(4mg - N = 4m\,a_{cm}\).
\(a_{cm} = \dfrac{(3/2 - 1/2)mg}{4m} = \dfrac{g}{4}\) downward. Then \(N = 4mg - 4m \times g/4 = 3mg\).
The centre of mass falls with acceleration \(g/4\), and the clamp holds the pulley with \(3mg\), less than the total weight \(4mg\). Check: the tension is \(T = 2 \times 3m \times m\,g/4m = 1.5mg\), and the clamp must hold \(2T = 3mg\).
The clamp force is less than the weight because the system is not in equilibrium. Its centre of mass accelerates downward, so the upward force must be smaller than the downward one. A spring balance holding the pulley would read less than the total weight while the blocks move.
Blocks of \(2\,\text{kg}\) and \(3\,\text{kg}\) hang over a light pulley clamped to the ceiling. Take \(g = 10\,\text{m/s}^2\). Find the force with which the clamp holds the pulley while the blocks move, in newtons.
Two blocks joined by a spring slide on a rough floor while a girl pulls the front block with a rope. The system is the two blocks and the spring. Which of these forces are external?
A person sitting in a stalled car pushes hard on the dashboard. Explain, using the centre of mass, why the car does not move forward.
Take the person and the car as the system. The push on the dashboard and the push of the seat back on the person are internal forces, a third-law pair that cancels in \(M\vec a_{cm} = \vec F_{ext}\). No new external force acts, so the centre of mass of the car and the person keeps its velocity, which is zero. The car cannot move forward on its own internal forces.
In the Principia of 1687, right after his three laws of motion, Isaac Newton listed several corollaries. The fourth concerns the common centre of gravity of two or more bodies. The actions of the bodies among themselves do not change its state of motion or rest. Newton saw that his third law, action and reaction, made the centre of a system obey the first law. That single remark lets us treat a planet, a star system or a spinning bat as one particle.
- Including internal forces, such as a spring force or a tension, in \(M a_{cm} = F_{ext}\).Draw the boundary of the system first. Forces between parts inside it cancel in pairs.
- Dividing the external force by the mass of one part.The centre of mass accelerates with \(F_{ext}\) divided by the total mass \(M\).
- Concluding that internal forces change nothing.They change how each part moves. They cannot change how the centre of mass moves.