On Diwali night a sky shot climbs from a rooftop in a gentle curve and bursts near the top of its flight. Hundreds of glowing sparks fly outward in every direction and form a growing ball of light. Watch the ball as it grows. Its centre does not stop at the burst. It drifts on across the sky and sinks, as if the unburst shell were still flying.
The centre of mass keeps its velocity
The last lesson proved \(M\vec a_{cm} = \vec F_{ext}\). Now suppose the net external force on a system is zero. Then \(\vec a_{cm} = 0\), and the velocity of the centre of mass cannot change. The parts may push, pull, walk or explode. Whatever they do to one another, the centre of mass moves in a straight line at a steady speed. A system with no net external force is called an isolated system.
If the net external force on a system is zero, the velocity of its centre of mass stays constant. In particular, if the centre of mass is at rest, it stays at rest.
The case of a system at rest is the most useful. If the centre of mass starts at rest, it stays at the same point. For two bodies that then move, \(m_1 x_1 + m_2 x_2\) cannot change, so their displacements obey \(m_1\Delta x_1 + m_2\Delta x_2 = 0\). One body moves forward only if the other moves back, by distances in the inverse ratio of the masses.
A man on a boat
A man stands at one end of a flat boat that floats at rest on still water. He walks to the other end. Water resistance on a slow boat is tiny, so there is no horizontal external force on the man and the boat together. His feet push the boat backward, and the boat pushes him forward. These are internal forces.
The man walks \(4\,\text{m}\) relative to the boat, as Fig 6.1 shows. The boat slides back a distance \(b\), so the man moves forward \(4 - b\) relative to the water. The centre of mass stays put: \(60(4 - b) = 140\,b\), so \(b = 240/200 = 1.2\,\text{m}\). The man moves \(2.8\,\text{m}\) forward and the boat \(1.2\,\text{m}\) back. When he stops, the boat stops too.
The walk of \(4\,\text{m}\) is measured along the boat, but \(m_1\Delta x_1 + m_2\Delta x_2 = 0\) needs displacements measured from the ground. Write the man's displacement from the ground as his walk minus the boat's backward slide. Using \(4\,\text{m}\) directly is the most common error in these problems.
A \(20\,\text{kg}\) dog stands on a flat \(80\,\text{kg}\) boat at rest on still water, \(15\,\text{m}\) from the shore. The dog walks \(5\,\text{m}\) along the boat towards the shore and stops. Ignore water resistance. How far from the shore is the dog now?
Take the system as the dog and the boat. No horizontal external force acts, and the system starts at rest, so its centre of mass stays fixed. Let the dog move \(d\) towards the shore and the boat \(b\) away from it, both from the ground.
The walk along the boat is \(d + b = 5\,\text{m}\). The centre of mass stays fixed: \(20\,d = 80\,b\).
\(d = 4b\), so \(5b = 5\) and \(b = 1\,\text{m}\), \(d = 4\,\text{m}\).
The dog has moved \(4\,\text{m}\) towards the shore, so it is \(15 - 4 = 11\,\text{m}\) from the shore, not \(10\,\text{m}\). The boat has moved \(1\,\text{m}\) away.
A \(50\,\text{kg}\) boy stands at one end of a \(150\,\text{kg}\) boat, \(6\,\text{m}\) long, at rest on still water. He walks to the other end. Ignore water resistance. How far does the boat move, in metres?
Two skaters and a rope
Two skaters stand at rest on smooth ice and hold the ends of a rope. When one or both of them pull, the tension pulls each towards the other with the same force \(T\). The tension is internal, and the ice is smooth. So the centre of mass of the pair stays where it was, and they meet exactly there.
The centre of mass is \(60 \times 10/100 = 6\,\text{m}\) from A, as Fig 6.2 shows. So A slides \(6\,\text{m}\) and B slides \(4\,\text{m}\) before they meet. It makes no difference who pulls. If only A pulls the rope hand over hand, the tension still acts on both, and they still meet at C. The lighter skater always covers more ground.
A \(30\,\text{kg}\) boy and a \(70\,\text{kg}\) man stand \(20\,\text{m}\) apart on smooth ice and pull on a rope between them. Only the man pulls. Where do they meet?
Why walking needs the ground
Now reverse the question. How does anything with only internal forces ever move its centre of mass? It cannot. To walk across a floor, you push the floor backward with your foot, and the floor's friction pushes you forward. That friction is an external force on you. On perfectly smooth ice, with no friction at all, you could swing your legs all day and your centre of mass would not move.
The same holds for a car. The engine and the gears are internal. They turn the wheels, and the turning wheels push back on the road. The road's friction on the tyres is the external force that accelerates the car's centre of mass. On an oily patch the engine roars and the wheels spin, but the car stays where it is.
A girl sits at rest in the middle of a perfectly smooth frozen lake, holding her school bag. Explain how she can reach the shore, using the centre of mass of the girl and the bag.
With no friction, no horizontal external force acts on the girl and the bag, so their centre of mass stays at rest where it is. She cannot move her own body towards the shore by any motion of her limbs. But she can throw the bag away from the shore. The bag then moves one way, and she must move the other way so that their centre of mass stays fixed. She slides towards the shore.
A shell that bursts in flight
A shell fired from a gun flies along a parabola, because gravity is the only external force on it. At some point a charge inside it explodes, and the shell breaks into fragments that fly in different directions. The explosion is violent, but every force in it acts between parts of the shell. Those forces are internal, and they cancel in pairs.
After the burst, the only external force on the fragments is still gravity, the sum of their weights, \(Mg\). So the centre of mass of the fragments has acceleration \(g\) downward, exactly as before. It continues along the same parabola that the unburst shell would have followed. This is a statement about the point C, not about any fragment.
Play the three chips of Fig 6.3. In the first, the shell bursts into equal halves, and one half is left at rest at the top. It falls straight down. The other half must go on so that C stays on the curve, and it lands at \(60\,\text{m}\). Both land together, and C lands at \(40\,\text{m}\), the range of the unburst shell. In the second chip, one third of the shell is thrown back along its path to the gun. The other two thirds land at \(60\,\text{m}\), and again C lands at \(40\,\text{m}\).
The third chip shows the limit of the rule. The halves are thrown upward and downward. The lower half reaches the ground first, at \(t = 3.24\,\text{s}\). From that instant the ground pushes on it, and the ground's force is external. The centre of mass leaves the parabola, and it finally rests at \(42.4\,\text{m}\), not at \(40\,\text{m}\).
The rule "the centre of mass of the fragments lands at the original range" needs all the fragments to land at the same moment. That happens, for example, after a burst at the top of the flight in which no fragment is thrown up or down. If one fragment lands first, the ground's force acts on it, and the rule no longer applies.
A shell would land \(120\,\text{m}\) from the gun on level ground. At the top of its flight it bursts into two equal halves. One half falls straight down from rest at the burst point. Ignore air resistance. Where does the other half land?
The half at rest at the top falls freely from the top, like the unburst shell's vertical motion. The vertical position of C also falls freely from the top. So the other half must fall at the same rate, and both halves land together.
The top of the flight is halfway along the range, at \(60\,\text{m}\). So the first half lands at \(x_1 = 60\,\text{m}\).
C lands where the unburst shell would have, at \(120\,\text{m}\). For equal halves, \(120 = (60 + x_2)/2\), so \(x_2 = 180\,\text{m}\) from the gun.
A shell fired on level ground would have a range of \(80\,\text{m}\). At the top of its flight it bursts into two fragments of masses in the ratio \(1 : 3\). The lighter fragment falls straight down from rest at the burst point, and both fragments land together. How far from the gun does the heavier fragment land, in metres?
A shell bursts in mid-air into two fragments. One fragment reaches the ground well before the other. After that moment, does the centre of mass of the two fragments stay on the original parabola?
Fifty passengers, each of about \(60\,\text{kg}\), walk \(10\,\text{m}\) from the back to the front of a ferry of mass \(12\,000\,\text{kg}\) moored loosely on still water. Ignore water resistance. Estimate how far the ferry moves backward, in metres.
Anyone who has stepped from a small country boat onto the steps of a ghat on the Ganga knows the danger. As you push off with your feet, the boat slides away from the steps. Your step forward carries you a shorter distance than you intended, and the boatman quickly holds the boat against the ghat with his pole. The pole brings in an external force from the ground, which the water alone cannot give.
Return to the Diwali sky shot of the opening. The burst is an internal event, so the centre of mass of all the sparks continues along the curve that the shell was following. The ball of light grows about that moving centre. Air drag slows the tiny sparks quickly, so in a real sky the ball soon stops drifting and sinks. That drag is an external force, and it ends the rule.
- Measuring a walk along a boat as the walker's displacement from the ground.The walk along the boat is the walker's ground displacement plus the boat's slide. Use ground displacements in \(m_1\Delta x_1 + m_2\Delta x_2 = 0\).
- Assuming that C lands at the original range after any burst.That holds only while all the fragments are in the air. Once one fragment lands, the ground's force is external.
- Thinking that the one who pulls the rope moves less.The tension pulls both ends equally; the distances depend only on the masses.