Place a steel katori on a smooth glass table and drop a marble in from the rim. The marble races down one side and up the other. The katori does not stay still. It shuffles back and forth under the marble, out of step with it. On a rough wooden table, the same katori hardly moves.
Two laws at once
Many problems put a small body on or inside a larger body that is free to slide on a smooth floor. The external forces are gravity and the push of the floor, and both are vertical. So the horizontal momentum of the pair is conserved. The vertical momentum is not, because the floor pushes harder or softer as the small body speeds up or slows down.
One conservation law gives one equation, but two speeds are unknown. The second equation comes from energy. The contact between the two bodies is smooth, so the normal force between them does no net work on the pair. Mechanical energy is then conserved too. With both laws, and the geometry of the contact, every speed follows.
The normal force between the block and the wedge does work on each body separately. It does negative work on the sliding block and positive work on the wedge. The two works cancel, because the contact is smooth and the two surfaces stay together. So the mechanical energy of the pair is conserved, but not the energy of either body alone.
A \(1\,\text{kg}\) block slides from rest down the smooth \(45°\) face of a \(3\,\text{kg}\) wedge. The wedge rests on a smooth floor. When the block has dropped \(1.4\,\text{m}\), find the speed of the wedge and the speed of the block. Take \(g = 10\,\text{m/s}^2\).
Let the wedge move back at \(V\). Let the block have horizontal velocity \(v_x\) forward and vertical velocity \(v_y\) downward, over the floor.
Momentum along x: \(1 \times v_x = 3V\). Relative to the wedge the block moves along the \(45°\) face, so \(v_y = v_x + V\). Energy: \(1 \times 10 \times 1.4 = \tfrac12 \times 1 \times (v_x^2 + v_y^2) + \tfrac12 \times 3 \times V^2\).
With \(v_x = 3V\) and \(v_y = 4V\), the energy equation is \(14 = 12.5V^2 + 1.5V^2 = 14V^2\). So \(V = 1\,\text{m/s}\), \(v_x = 3\,\text{m/s}\) and \(v_y = 4\,\text{m/s}\).
The wedge moves back at \(1\,\text{m/s}\), and the block moves at \(\sqrt{9 + 16} = 5\,\text{m/s}\). On a fixed wedge the block would have reached \(\sqrt{2 \times 10 \times 1.4} \approx 5.29\,\text{m/s}\).
A block slides down a smooth wedge that rests on a smooth floor. Which statement about the block and the wedge together is right?
A groove in a free block
Now cut a smooth semicircular groove of radius \(a\) into a block A, and rest it on a smooth floor. Hold a small block B at the rim of the groove, where the radius is horizontal, and let it go. B slides down, across the bottom and up the far side. All the while the groove block moves the other way.
Block B, of \(1\,\text{kg}\), is released at the left rim of the smooth groove in block A, of \(4\,\text{kg}\). The floor is smooth. What happens when B reaches the far side of the groove?
Play the figure above. The two momentum arrows are always equal and opposite, and the readout stays at zero. At the bottom of the groove, B moves at \(4\,\text{m/s}\) and A at \(1\,\text{m/s}\) the other way. By then A has moved back \(0.2\,\text{m}\). At the far rim both stop, and A has moved back \(0.4\,\text{m}\) in all.
In the figure above, block A has mass \(M = 4\,\text{kg}\) and a groove of radius \(a = 1\,\text{m}\). Block B has mass \(m = 1\,\text{kg}\) and is released at the rim. The floor and the groove are smooth. Take \(g = 10\,\text{m/s}^2\). Find the speeds of A and B when B is at the bottom, and how far A has moved by then.
At the bottom both velocities are horizontal. Let B move at \(v_1\) and A at \(v_2\) the other way. Momentum: \(1 \times v_1 = 4\,v_2\), so \(v_1 = 4v_2\).
B has dropped \(a = 1\,\text{m}\). Energy: \(1 \times 10 \times 1 = \tfrac12 \times 1 \times v_1^2 + \tfrac12 \times 4 \times v_2^2 = 8v_2^2 + 2v_2^2 = 10v_2^2\). So \(v_2 = 1\,\text{m/s}\) and \(v_1 = 4\,\text{m/s}\).
The horizontal position of the centre of mass cannot change, since the pair started at rest. B moves \(1\,\text{m}\) to the right relative to A. So A moves back by \(\dfrac{m}{m + M} \times 1 = 0.2\,\text{m}\), and B moves \(0.8\,\text{m}\) over the floor.
At the bottom, the force between A and B is vertical, so A is not accelerating at that instant. Its frame is then as good as an inertial frame, and B moves in a circle of radius \(1\,\text{m}\) relative to it at \(4 + 1 = 5\,\text{m/s}\). So the normal force on B there is \(mg + mv_{\text{rel}}^2/a = 10 + 25 = 35\,\text{N}\). A fixed groove would give \(10 + 20 = 30\,\text{N}\).
A \(6\,\text{kg}\) block on a smooth floor has a smooth semicircular groove of radius \(0.8\,\text{m}\). A \(2\,\text{kg}\) block is released at the rim of the groove. Take \(g = 10\,\text{m/s}^2\). Find the speed of the \(6\,\text{kg}\) block when the small block passes the bottom, in m/s.
In the example above, block B of \(1\,\text{kg}\) slides from the left rim to the right rim of the groove, of radius \(1\,\text{m}\), in block A of \(4\,\text{kg}\). How far has A moved over the floor by then, in metres?
A bead on a wire and a swinging bob
A bead of mass \(M\) slides freely on a smooth horizontal wire. A bob of mass \(m\) hangs from it on a light string of length \(L\). The bob is held against the wire with the string taut and is let go. The external forces are gravity and the push of the wire, both vertical. So the horizontal momentum stays zero, and the centre of mass never moves sideways.
This fixed line gives the distance at once. At the start the bob is \(L\) to the side of the bead, and C divides that gap in the ratio \(m : M\). When the string makes an angle \(\theta\) with the wire, the sideways gap is only \(L\cos\theta\). The gap has shrunk by \(L(1 - \cos\theta)\), and the bead's share of that shrink is \(m/(M + m)\).
For the figure, \(m/(M + m) = 1/3\) and \(L = 0.6\,\text{m}\), so \(d = 0.2(1 - \cos\theta)\). When the string hangs straight down, \(\theta = 90°\) and the bead has slid \(0.2\,\text{m}\). At that instant both velocities are horizontal. Momentum gives \(2v_1 = 1 \times v_2\) for the bead and the bob. Energy gives \(1 \times 10 \times 0.6 = \tfrac12 \times 2v_1^2 + \tfrac12 \times 1 \times v_2^2\), that is \(6 = 3v_1^2\). So the bead moves at \(\sqrt2 \approx 1.41\,\text{m/s}\) and the bob at \(2.83\,\text{m/s}\).
A \(3\,\text{kg}\) bead slides freely on a smooth horizontal wire. A \(1\,\text{kg}\) bob hangs from it on a string \(0.8\,\text{m}\) long, and is released with the string along the wire. How far has the bead slid when the string makes \(60°\) with the wire, in metres?
As the bob in Fig 4.1 swings down, how does the centre of mass C of the bead and the bob move?
Now return to the katori on the glass table. The glass is smooth, so the horizontal momentum of the bowl and the marble stays zero. When the marble runs right, the bowl runs left, and the centre of mass stays on one vertical line. With the two masses known, energy then gives the speeds, just as it did for the groove above. On a rough wooden table, friction from the table is an external force, and the bowl stays nearly still.
- Conserving the vertical momentum of a block on a free wedge.The floor pushes up with a changing force. Only the horizontal momentum is conserved.
- Using the speed \(\sqrt{2gh}\) of a fixed wedge or a fixed groove.The free block takes part of the energy. Solve momentum and energy together.
- Forgetting that the sliding block's velocity relative to the wedge lies along the face.Write the relative velocity, \(v_x + V\) horizontally, and use the slope of the face.