Undergraduate Mechanics Page - Some Thoughts, and Solutions
Ralph Shiell, Physics Department, Trent University, Ontario, Canada
Welcome. Here you will find some notes and solutions to the mechanics questions found here.
There are seven questions in total.
Question 1.
Consider (using Cartesian coordinates) vector A = (1, 1, 1). One example of A is the vector that points from the origin to the point x = 1, y = 1, z = 1.
But A is also (and equivalently) the vector that points from, say, (3, 8, 9) to position (4, 9, 10). Vector A has magnitude and direction only, period.
This physical "entity" with magnitude and direction might be a force acting on a mass that happens to be at a particular location. That is ok, but the "force" in itself has no position associated with it. If the point of application moves, then whatever has caused that force has done work on your mass. However it is not the force that moved, but instead the point of application of the force.
Consider the angular momentum about the origin of a 1 kg particle moving in a circle in the xy plane in an anticlockwise direction looking down the z-axis, with radius 1 m and a speed of 1 m/s. The angular momentum about the origin, LO, is given by r × p, where r is the vector from the origin to the particle and p the linear momentum of the particle. LO is therefore the vector 1 k. This physical quantity (angular momentum of the particle about O) has magnitude (value 1) and a direction (along z) but it has no position. You could put this vector pointing away from the origin - a sensible choice in a way, as that is the point about which you are considering the angular momentum, but you do not have to. The angular momentum itself is "of no fixed abode".
So the correct answers are "b" and "c".
Question 2.
Crucial to solving all mechanics problems is determining what your physical "system" actually is. You can choose whatever system you wish, and from this determine the behaviour of that system by considering the presence (or lack of) external interactions (i.e. forces due to "stuff" outside your system).
Take an example of a book dropped to the floor. If we are interested in the fate of the book, our system might be the book. The earth is then outside this system and it exerts a force on the system. This force changes the linear momentum of the book (according to Newton's second law), and because the point of application of this force has moved along the direction of this force, the force of gravity has done work on the book and so changed the total energy of the book.
Has the angular momentum of the book changed? Well, at present that's an incomplete question. We have not said about which point we wish to consider the angular momentum. What we do know is that
.
Note we have to consider the same point, A, on both sides of this equation }
So if point A is such that the force of gravity acting on our book produces a torque about this point, then yes, the angular momentum of the book about point A will have changed. Therefore as the book falls the angular momentum about any point not directly below or above the book will therefore have changed .
If, instead, we had considered the book and the Earth to be our system and we had ignored any matter outside this system then the linear momentum of this system is constant (the book falls towards the Earth and the Earth falls towards the book), the total energy of the system is constant (the book and Earth gain kinetic energy at the expense of potential energy) and the angular momentum of the system about any point in space is constant (if it was rotating, it continues to rotate).
So the correct answer is "d".
Question 3.
The "value" of a scalar is its magnitude. The "value" of a vector is a set of three numbers (its three components) that tell you both its magnitude and its direction. We cannot say that a vector is bigger than a scalar, any more than we can say that a vector equals a scalar (and we can't!). They are very different entities - but both very useful (essential) to describe the physical world.
Answer (a) is a time derivative of a vector, and is the vector (1, 2, 0). This is the velocity of the particle (in ms-1).
Answer (b) is the rate of change of the distance of the particle from the origin. This distance is (5t2+4)1/2 and so the rate of change of this distance at t = 1 sec is 5/3 ms-1.
Answer (c) is the magnitude of the vector in (a) - i.e. the magnitude of the velocity. This is 
ms-1, and is the speed of the particle.
So the correct answer is "d".
Question 4.
We can consider the body to be made up from a pile of different sized disks as shown. The axis of rotation for each disk is, in general, off - centre.
Note that if this rigid body is not allowed to deform then these axes of rotation point in the same direction for all disks. Further, each disk must rotate at the same rate - or one part of the rigid body would catch up with another. Putting this together, each point within the body has the same angular velocity vector, ω, (same magnitude and direction) and each point follows the path of a circle, with velocity vector determined by v = ω× r, where r is a vector from any point on the axis of rotation to the point of interest. It is worth taking a moment to check the right - hand rule for cross products, as cross products are ubiquitous in physics (here, if ω is pointing up, then the part of the cylinder on the right hand side of the figure is moving into the page).
[In fact a particle following any arbitrary curve can always be considered at a given instant to be moving in a plane circular path about a certain axis. The (instantaneous) plane of rotation is defined by two consecutive non - collinear path elements, ds1 and ds2. If we let this part of the curve lie in the xy plane, according to y=f(x), then the instantaneous axis of rotation is the line normal to this plane (which then lies parallel/antiparallel to the z axis), at a distance
.
The only problem with an arbitrary (i.e. non - circular) curve is that the axis of rotation changes direction and position constantly, and the angular velocity vector, ω, changes direction and magnitude too. However, if all points on a body follow a path ds1, ds2 and ds3 etc. that all lie in a plane such that all points maintain the same position relative to each other, then we have an axis of rotation that is fixed relative to the body and is the same for all points within the body. For steady rotation the magnitude of ω is constant and the speed of any point is constant in time.]
So the correct answer is "a".
Question 5.
The angular momentum (which must be defined about a point to be a meaningful quantity*) of a system of particles is given by the vector sum of all the angular momenta of each particle within the system. (It is a perfectly good vector quantity, after all, and vectors that represent the same physical quantity add). The angular momentum of each particle about point A is LA = r × p, where r is the vector from point A to the particle, and p is the particle's linear momentum.
Here we consider one particle only, and so we calculate r × p at various points in its trajectory. Now the vector r × p has a magnitude given by the area of the parallelogram determined with sides of length r and p (this area is its base × perpendicular height). The perpendicular height is always the same (it is the y value of the particle) and the length of the base is always the same (it is the linear momentum). So the angular momentum of particle A about any point fixed in space takes the same value at all times.
Alternatively, we could have invoked
.
The particle is moving in a straight line at constant speed, therefore there is no net force on the particle. There is therefore no torque on the particle about any point. And therefore the angular momentum of the particle about any point (including A) is constant.
So the correct answer is "a"
[*It turns out that the angular momentum of any system of particles about a point A is the angular momentum about point A of a particle with the total mass of the system following the path of the centre of mass of the system plus the angular momentum of the body about its centre of mass position.
In equation form: LA = Lcom + rcom × Ptot
, where rcom is the vector from point A to the centre of mass and Ptot is the total linear momentum of the system.
The centre of mass is thus a special point for a body. Sometimes, therefore, when we talk about the angular momentum of a body, we mean "about the centre of mass". However, if you are not told this explicitly, you are fully entitled to ask for clarification.]
Question 6.
We shall consider our "system" of particles to be the electron and proton, which are assumed isolated from any other matter. The mass of the proton is so much greater than that of the electron that we shall assume the proton remains at rest at all times. The total charge of this system is zero, and a hydrogen atom does contain an electron and proton. So it all looks ok ...
However, let us consider the potential energy of the system and the total mechanical energy of the system. We shall plot these different quantities on the same graph. They have the same units (that of energy), so we need use only one y axis. The potential energy of this system is given by a -1/|r|+c shaped curve, where r is the distance between the particles. c is an arbitrary constant - this can take any value and we still have the correct physics. It proves convenient to set c=0 - and so the potential energy is zero at large |r|. Then at large r, where the potential energy is close to zero, the only energy of the system is due to the kinetic energy of the electron.
We have plotted the potential energy in blue, and the total mechanical energy in red. Because our system is isolated, we know that the total energy of the system is conserved. As the electron moves towards the proton the potential energy of the system decreases, and so the kinetic energy must increase. The difficulty with us making a hydrogen atom is that this atom is made up of a bound (electron - proton) system - i.e. one in which the particles have insufficient energy to escape from each other. So a hydrogen atom corresponds to a total mechanical energy less than zero on our plot.
We can see that in fact the electron will accelerate towards the proton, pass very close to it at a breakneck speed, and come away again. At a large distance from the proton, it will still have a large kinetic energy (whatever it started with). So to make a hydrogen atom from a separated electron and proton, we need something to take away some energy.
So the correct answer is "b".
[If you are surprised by this, don't worry - it is not obvious. In our everyday world we are so used to the presence of friction that we can find it difficult to conceive a world without it. However, to understand physics, we first need to understand interactions between particles in such a world. We can then add friction (which is simply another set of microscopic interactions that obey the same laws of physics) and then we do get the results from everyday experience. It is because the laws of physics are so transparent in a world without friction that we often discuss experiments involving
microscopic particles (as in this question), or collisions between hockey pucks on flat ice rinks (as below), or the heavenly bodies (moon, planets, stars, etc.)
If, instead, our electron and proton were oppositely-charged ball bearings rolling on a flat rubber mat, then the total energy of the two-ball-bearing-system is not conserved (now there is friction due to the mat). As one ball bearing
approaches the other the total energy of our system can become less than the potential energy at a large distance and the ball bearings will indeed become "bound" to each other. Mechanical energy has been lost from our (simplified) two-body system and gone into heating of the ball bearings and of the mat.]
Question 7.
We can take our "system" to be the moving puck and the dumbbell. Assuming this to take place on a flat ice rink where there is no friction, and no net external force on any part of the system (there are in fact two external forces on each mass - their weight and the reaction force due to the ice, but these cancel each other exactly). The total linear momentum of the system and total angular momentum of the system about any point you care to pick are therefore conserved. Is the energy conserved? No net external force on the system certainly means the total energy is conserved, but not necessarily the total mechanical energy (i.e kinetic plus potential energy of the macroscopic bodies). In fact, we are told that the puck collides inelastically, and so from Question 6 above, we know that some mechanical energy must have been lost to heat (of the glue and the pucks).
We can use the conservation of linear momentum of the total system. Before the collision we have a system with a particular linear momentum, pointing along the x axis with a particular magnitude. I have demonstrated this by denoting the motion of the centre of mass by the motion of the blue cross in the picture below, assuming the pucks have equal mass, and the collision hits "head-on".
After the collision we must have the same total linear momentum. So the centre of mass of the resulting dumbbell must continue to move in its original direction. And as the centre of mass was moving parallel to the x axis and above the x axis, it continues to do so for all time.
We can use the conservation of angular momentum about any point we wish. The simplest point to choose might be a point on the x axis. Before the collision we have a system with zero angular momentum about this point - i.e. the vector (0, 0, 0). So after the collision we must have the same angular momentum about this point. But we know that the linear motion of the resulting dumbbell is to the left with its centre of mass above the x axis, which constitutes a non - zero angular momentum vector pointing out of the page. This must be exactly balanced by a non - zero angular momentum vector pointing into the page, and this can be achieved if the resulting dumbbell moves in a clockwise manner when seen from above:
So the correct answer is "a" and has been obtained simply by considering the conservation of linear momentum and conservation of angular momentum of the system (about any point we wish). This answer is still correct if the pucks have different masses, if the rod has some mass, or even if the moving puck does not hit the bottom puck "head-on". Pretty good, don't you think?
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