FURTHER MECHANICS - for 2011-12 students

New: videos on SHM:

Part 1, explaining the idea of SHM and a couple of very simple examples.

Part 2 - examples from EdExcel M3, showing typical techniques for calculating key quantities.

Part 3 - 2 exam questions (fairly difficult) involving how to find approximate SHM equations

The sequence of topics is as follows:

• Projectiles (M2 topic)
• Motion in a circle with constant speed (M2 topic)
• Motion in a vertical circle (FM topic)
• Equilibrium of a Rigid body (FM topic)
• Momentum and Impulse (FM topic)
• Hooke's Law (M2 topic)
• Simple Harmonic Motion (FM topic)
• Rotation of a Rigid Body (FM topic)

A little note: the last two topics are definitely the most difficult so save up some brain power for Semester 2!!

Some notes on how to do rigid body equilibrium problems:

1. Have confidence! The information must be there, available, to solve the problem. Since the only physical laws we apply are the laws of static equilibrium: total force is zero, total moment is zero, and the laws of static friction, there are only two types of equation + plus one type of inequality that can possibly apply:
$\sum_i \textbf{F}_i=\textbf{0} \sum_i \textbf{F}_i \times d_i = 0 f_i \le \mu R_i \quad \textbf{for each friction force}\ f_i$

This means that the problems cannot be complicated. In almost all cases, you will only need to solve linear simultaneous equations; calculus or series are not required. Please do remember to use inequalities rather than equations for the friction forces - this is probably the most common mistake.

2. Search for a pivot.
The principle here is that, in static equilibrium the sum of the clockwise moments about ANY PIVOT INSIDE OR OUTSIDE THE BODY is zero. Since the moment of a force depends on the perpendicular distance from the pivot to the line of action of the force, you are looking for a pivot through which as many as possible of the lines of action of the individual forces pass. Pay attention to corners, centres of bodies and in general to any point which has a high degree of symmetry on the diagram. You need to learn this skill yourself, the examiner will rarely if ever tell you what to do!

3. Make a good free body diagram of the body you're analysing.
In these questions, the examiners will almost always provide you with a diagram. Sometimes they mark on the forces themselves, other times they don't - you can't rely on it. In any cases, it is 100% hopeless to try and solve problems of this complexity without a good quality diagram to look at, so make at least one good quality diagram with a ruler which marks all forces acting on the body you're analysing. It is quite common that you will need to draw more than one diagram - such as two free body diagrams of two connected objects, or sometimes make a diagram of one section of the system to help you analyse the geometry correctly.

4. Unambiguous labelling is a necessity.
It doesn't matter if you choose the letter N or the letter R to represent normal contact forces, but what is EXTREMELY important is to make clear which force you are talking about in your equations. Use subscripts like this:

$F_1 \le \mu N_1 \quad \quad F_2 \le \mu N_2 \quad N_1 +F_2=X$

and of course just as important is that these labels match what's on your diagram, otherwise the examiner cannot follow your arguments.

5. Remember the main property of the frictional and normal contact forces.
Friction between two surfaces opposes the relative motion between the surfaces of contact. This will tell you the direction of the force. The normal contact force is always perpendicular to the surface of contact. If the surface is not flat, the direction of the contact will be in the direction of the normal line at the point of contact. The normal contact force between flat surfaces will not always act through the centre of mass of the object concerned - see 6.

6. Toppling and Lami's Theorem.
It can be proved (you don't need to learn how if you're not concerned) that if exactly three forces act on a body, then it will be in equilibrium (neither move nor rotate) if and only if (a) the three forces as vectors form a complete triangle and (b) the line of action of the three forces all pass through the same point.
When trying to find the point at which an object, resting on a surface, is about to topple we can use Lami's Theorem. The lines of action of the normal contact force and the weight must cross at the surface of contact. This will be possible as long as the line of action of the weight (which is a vertical line through the centre of gravity, by definition) passes through the joint surface of contact. The normal contact force can (effectively) act through any point on the contact surface (it is best to think of it as a weighted average of lots of little contact forces between all the atoms in contact). Once this point crosses outside the surface of contact, static equilibrium is impossible and the body will rotate. This is called toppling.

In more complicated, more general cases, it will still be true that on the point of toppling (just about to topple), the normal contact force will act at the corner which is the pivot of the potential rotation.

ROTATIONAL DYNAMICS

Start by considering motion around a circle. We know from our studies that a particle rotating in a circle has motion which obeys these equations:

$a_c = \frac{v^2}{r}$

$v = r \dot{\theta}$

and we also remember that, in these equations the velocity v is a vector tangential to the circle, whilst the acceleration a is centripetal which means that it's directed along the radius towards the centre. The quantity

$\dot{\theta}$

is referred to as the angular velocity. The dot above theta means the first derivative with respect to time. Let's introduce further notation:

$\dot{\theta} = \frac{\textrm{d}\theta}{\textrm{d}t} = \omega$

$\ddot{\theta} = \frac{ \textrm{d}^2 \theta } { \textrm{d}t^2 } = \alpha$

The second of the two equations refer to what's called the "angular acceleration", which in some books is written as "alpha", or "theta double dot" more commonly.
How can we apply these quantities to the case of circular motion with varying speed (the most common example is motion in a vertical circle, but we will have to consider others also)?

Careful consideration should show you:

$a_t = r \ddot{\theta}$

$a_c=r\left(\dot{\theta}\right)^2$

These are the two components of the acceleration of the particle; the first, the transverse acceleration, is the product of the radius with the angular acceleration. Notice that you can find the magnitude of he acceleration of the particle by combining the two perpendicular components:

$|\underline{\textbb{a}}|^2 = a_t^2 + a_c^2 = \left(r\ddot{\theta}\right)^2 + \left(r\left(\dot{\theta}\right)^2\right)^2$

Since these equations describe the motion of any particle moving in a circle, they can also be used to apply to the rotation of a rigid body, because each particle in the body rotates in a circle with the same angular velocity, precisely because it's rigid. The body will obey Newton's laws, of course, and it's possible to make a rotational analogue of all the equations we learnt in M1 for linear motion. The analogy is as follows:

 Linear Rotational x $\theta$ u, v $\dot{\theta}$ or angular velocity a $\ddot{\theta}$ F $\tau$ or "torque" or"moment" or "turning effect" m $I$ or "moment of inertia"

Using these analogies, all the linear equations of motion with constant acceleration can be recreated and are correct:

 Linear Rotation $F=ma$ $\tau = I \ddot{\theta}$ $v = u +at$ $\dot{\theta}_f = \dot{\theta}_i + \ddot{\theta}t$ $v^2=u^2+2as$ $\dot{\theta}^2_f = \dot{\theta}^2_i+2\ddot{\theta}\theta$ $s=ut+\frac{1}{2}at^2$ $\theta = \dot{\theta}_i t + \frac{1}{2}\ddot{\theta}t^2$ $E_k = \frac{1}{2}mv^2$ $E_k = \frac{1}{2}I\dot{\theta}^2$

Of these, the first and last are the equations that you will use most frequently. The first one allows you to find the angular acceleration for a given "torque" (or moment). The last can be used to find the final angular velocity after a period of motion during which the torque, and therefore the angular acceleration are not constant (problems of this type are common on the exam).

Finding the moment of inertia.
The moment of inertia is the rotational equivalent of mass. Just as the force needed to create 1 metre per second of acceleration is proportional to the mass, so the torque needed to create 1 radian per second squared of angular acceleration is proportional to the moment of inertia. It's easy to see that the moment of inertia of a single particle rotating at a distance r from a pivot should be mr^2:

$F = ma \Rightarrow \tau = Fr = mra = mr(r\ddot{\theta})=mr^2\ddot{\theta} = I\ddot{\theta}$

And we can find the total moment of inertia of a rigid body X can be found as a sum:

$I_X = \sum_i m_i r_i^2$

For continuous bodies of two or three dimensions such as triangular or rectangular laminas, rods, cubes, spheres or cones, we can use integration to accurately find the correct value of these sums. Key results are given in the formula sheet for Further Maths, MF10. NOTE THAT YOU WILL NOT BE EXPECTED TO FIND THE MOMENT OF INERTIA BY INTEGRATION, although it is discussed in your textbooks.

Finding related moments of inertia.
There are three rules which can be used to find moments of inertia of objects that are in some way similar to one given as a standard result. The stretch theorem (not often used on the exam) is the easiest one to understand.
The stretch theorem is the trivial observation that the moment of inertia of a 3D body which can be formed by stretching a 2D body along the axis of rotation has the same moment of inertia as the 2D body. For example, the moment of inertia of a circular disc of mass m and radius r is:

$\frac{1}{2}mr^2$

The moment of inertia of a cylinder, about its axis of symmetry has exactly the same formula.

The parallel axes theorem, which is a little less obvious, but still quite easy to prove, states that the moment of inertia of a body X about an axis A is equal to the moment of inertia of the same body X about an axis B which passes through the centre of mass G of X, plus the mass of X multiplied by the square of the distance between A and B:

$I_{A,X} = I_{B,X}+mr_{AB}^2$

To understand this rule, it helps to remember that it is logical that the moment of inertia will always be lower for an axis through the centre of mass than any other parallel axis - this is because in some sense the mass is more "evenly distributed" about the centre of mass than about any other point. Notice that the word "moment" is also used in statistics - the variance of a distribution is the "second moment" of the distribution, and is analogous to the moment of inertia.

The perpendicular axis theorem is the most complicated of the three, and is a bit more difficult to prove - but you don't need to remember the proof. It states that for any rigid lamina (2D body),

$I_z = I_x + I_y$
where Ix and Iy are moments of inertia about any two perpendicular axes in the plane of the lamina, and Iz is the moment of inertia about the axis which is perpendicular to the lamina. Very often you will be able to make use of this theorem by combining it with some symmetry of the body you are analysing. For example, the moment of inertia about the diagonals of a square of side a:

can be found by noting that:
1. According to the list of formulae, the moment of inertia of the square about an axis perpendicular to its plane and through its centre is:

$I_1 = \frac{1}{3}m\left(\left(\frac{a}{2}\right)^2+\left(\frac{a}{2}\right)^2\right)$

so that the moment of inertia for a square is:

$I_1 = \frac{1}{6}ma^2$

2. The moment of inertia I about each diagonal must be equal by symmetry.

3. According to the perpendicular axis theorem, therefore:

$I_1 = I + I = 2I$

From these three points we can deduce:

$I = \frac{1}{12}ma^2$

Finding the moment of inertia of a lamina (a two dimensional object of uniform mass per square metre) is one of the most common simple questions on the exam and is a source of a lot of easy marks, as long as you understand the few simple ideas above. The ideas you need to combine are:
• For a uniform lamina/body, the mass of one section of the body is proportional to the area/volume
• The moment of inertia of two or more bodies about the same axis is the sum of the individual moments of inertia
• The moment of an inertia of a lamina with a piece cut out can be found by subtraction (trivial application of the principle above)
• Standard formulae for the moments of inertia of a particle, a rectangle, a disc, a rod
• The three theorems described above: stretch rule, parallel axis theorem, perpendicular axis theorem
• The deduction that the moment of inertia about certain different axes must be the same by symmetry

Combining these ideas to solve problems in rotational dynamics.

Here is one very typical example of how we can use these formulas to make deductions about rotational motion (it's adapted from the CIE exam in Winter 2008):

Question: "A uniform disc of mass m and radius a, is free to rotate without resistance in a vertical plane about a horizontal axis through its centre. A light inextensible string has one end fixed to the rim of the disc, and is wrapped about the rim. A block of mass 2m is attached to the other end of the string. The system is released from rest with the block hanging vertically. While the block moves it experiences a constant resistance to motion of magnitude 0.1 mg. Find the angular acceleration of the disc and find also the angular speed of the disc when it has turned through one complete revolution."

As the mass falls, it is pulled upwards by the tension force in the string. This same tension force T is exerted on the disc, so that the net torque on the disc is Ta. To find T, we notice that the acceleration of the points on the circumference of the disc must be the same as the acceleration of the 2m mass. This allows us to use Newton's 2nd Law in linear form for the mass, and in rotational form for the disc, yielding 2 simultaneous equations to solve for T:

$2mg -T - \frac{1}{10}mg = 2m\ddot{x} \quad (1) \tau = Ta = I_{\textrm{disc}}\ddot{\theta} = \frac{1}{2}ma^2 \frac{\ddot{x}}{a} \quad (2) (2) \Rightarrow \quad \ddot{x} = \frac{2T}{m} \quad (3) (1) \Rightarrow \quad 2mg-T -\frac{1}{10}mg =4T \quad (4) (4) \Rightarrow 5T = \frac{19}{10}mg \quad (5) (5),(2) \Rightarrow \ddot{\theta} = \frac{\frac{19mg}{10}-\frac{19mg}{50}}{2ma} = \frac{19g}{25a}$

To find the angular velocity after one revolution, let's apply the principle of conservation of energy; remember that the disc acquiries a kinetic energy of rotation. "One revolution" means the angular displacement is 2pi.

$\textrm{Distance moved} = x = a\theta = 2\pi a$

Energy output = Final KE of disc + Final KE of mass + energy lost to resistance force

Energy input = GPE lost by 2m mass

$KE_{disc} = \frac{1}{2}I\dot{\theta}^2 = \frac{1}{4}ma^2\dot{\theta}^2 KE_{mass} = \frac{1}{2}(2m)v^2 = ma^2\dot{\theta}^2 \Delta GPE = (2m)g(2\pi a) = 4mg\pi a \Rightarrow 4mg\pi a = ma^2\dot{\theta}^2 + \frac{1}{4}ma^2\dot{\theta}^2 + \frac{1}{10}mg(2\pi a) \Rightarrow \dot{\theta} = \sqrt{\frac{76g\pi}{25a}}$

Answer this question: could we have found this answer without using the conservation of energy?

Yes. Because the angular acceleration of the disc is constant, we could use the rotational analogue of v^2=u^2 + 2as:

$\dot{\theta}^2 = 2\ddot{\theta} (2\pi)$
which, in this case, gives the same answer more quickly. But remember that in many questions the angular acceleration will not be constant because the torque is not constant. In these cases you'll have to use energy.

A quick conceptual summary about Simple Harmonic Motion

The following powerpoint describes many of the key concepts (although warning, the "Physics" version of the equations):

The set of notes you were given on paper also summarize the basics very well.
However what is not well explained anywhere in the books is the method of approximation to SHM. It works like this:
given an equation for net force which is complicated and certainly not linear, such as

$F = -f(x) \Rightarrow \frac{d^2x}{dt^2} = -\frac{a + bx + cx^2 + dx^3 + \ldots}{m}$

We can assume that this motion is approximately SHM if:
$a = 0 \Rightarrow \frac{d^2x}{dt^2} \simeq -\left(\frac{b}{m}\right)x$

How does this form arise? The answer is actually quite deep - as you learnt in elementary calculus, any curve looks like a line in a very small region. Another way to say this is that every function is "nearly" linear for small enough variations in the independent variable. That also is true for the function above representing the variation of force with distance. However if we want a form

$y=bx$

and not

$y=a+bx$
then it's necessary to choose the variable x such that y is zero when x is zero. That's why the important "trick" in analysing approximate SHM is to CHOOSE THE DISPLACEMENT VARIABLE SUCH THAT THE FORCE IS ZERO WHEN x is ZERO!! If you remember nothing else, please remember this!

Now how do we get:

$f(x) \equiv a + bx + cx^2 + dx^3 + \ldots$
? The answer is we can either use the method of Taylor Series (see p. 260 of yellow book for Further Pure) (but this is NOT on the syllabus), or we can use a binomial approximation:

$(1+x)^{n} \equiv 1 + nx + \frac{n(n-1)}{2!}x^{2} + \ldots$

SHM approximation for rotations
It's also possible to use the approximations

$\sin{\theta} \simeq \theta \Leftrightarrow |\theta| \ \textrm{small} \cos{\theta} \simeq 1 \Leftarrow |\theta| \ \textrm{small}$

to make an approximate form of an SHM equation; we convert the rotational analogue of Newton's 2nd Law:

$\tau = I \ddot{\theta} \tau = f(\sin{\theta},\cos{\theta}) \simeq k\theta \quad \Rightarrow \ddot{\theta} \simeq \frac{k\theta}{I}$

In this case the constant k must be negative for an SHM approximation, and the period can then be found in the normal way:

$T=\frac{2\pi}{n} = 2\pi\sqrt{\frac{I}{k}}$

SOME OTHER WORKSHEETS/FILES etc.

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Projectiles 2:

This worksheet was not issued in the lesson; you can ask for a copy from me or download it here.

Test - answers for projectiles test are available in this document: