CHAPTER 10

Rotations

You might not have realized it, but until now all the objects whose motion we studied (divers, cats, airplanes, etc.) were actually treated as dimensionless "point-like" objects. This is because, even though the objects had some non-zero size, we only considered translational motions, ignoring any possible rotation (a geometrical point, having no spatial extension cannot rotate around itself..). But now we are ready to analyze the more realistic situations that include both rotations and translations. To start, we remind ourselves of the "natural" unit for measuring angles, i.e. the radian: given an angle, the ratio between the arc it subtends and the radius of the arc is always the same, regardless of our choice for the radius. One then defines as the measure of the angle (in radians) the ratio between arc length s and radius r

$\theta$ (in radians) = s/r

Being the ratio between two lengths, radians are dimensionless. Converting radians to and from degrees is easily accomplished via

radian / degree = $2\pi /360^{0}$,

from which can can e.g. see that 1 radian = $360^{0}/2\pi = 57.3^0$

Rotational Variables

Even though in principle rotations can be described in terms of displacements, velocities and accelerations, it soon becomes obvious that these are not the most convenient quantities, for the following reason :

consider e.g. a rigid disk rotating around an axis going through its center; in a given time interval, a point at the disk's edge covers a longer distance than a point nearer the center : for a rotating body, velocity is not a well defined quantity, since it depends on the location of the point under exam. On the other side, in a given time interval, the angle swept by a radius joining the center of rotation with any arbitrary point is the same, regardless of the point's location. Rotations can then be better described in terms of angular, rather than linear, variables. We then define :

• angular displacement $\Delta\theta = \theta_{2}-\theta_{1}$ : the angle swept by a radius of a rotating body, i.e. the angle swept by a segment going from the center of rotation to an arbitrary point on the body.
  
  By convention, clockwise rotations correspond the negative displacements (why? because, when rotating clockwise, a given radius moves from larger to smaller angles...).
• average and instantaneous angular speed $\omega_{av} : \Delta\theta / \Delta t , \omega = lim_{\Delta t\rightarrow 0}\Delta\theta /\Delta t = d\theta /dt$ Angular velocities are best measured in radians/sec, dimension = T^-1
  
  Often angular velocities are expressed in terms of revolutions per second (or even revolutions per minute, rpm). How is this related to radians/sec?
  
  
  rev/sec = rev/sec x rad/rad = rad/sec x rev/rad = $\omega_{rad/sec}\times 1/(2\pi )$
  If the rotation occurs at a constant rate, we can define the period T as the time to perform one full rotation, and consequently
  $\omega_{rev/sec} = 1/T = \omega_{rad/sec}/(2\pi)$
  Notice also that $\omega_{rev/sec}$ is the frequency f, with f = 1/T.
• average and instantaneous angular acceleration:
  $\alpha_{av}= \Delta\omega /\Delta t, \alpha = lim_{\Delta t\rightarrow 0}\Delta\omega /\Delta t = d\omega /dt$
  
  .

Are angular velocities scalars or vectors? This is equivalent to asking : can we associate a given rotation with a direction in space? The answer is yes : a rotation is identified by the axis of rotation. We then adopt the convention of assigning to the angular velocity vector $\vec{\omega}$ the direction of the rotation axis. The vector orientation will be chosen so that, if you lie along the axis, the vector going from your feet to your head, then the rotation appears to be counterclockwise (see also right hand rule).

Once we define $\vec{\omega}$ as a vector, then also $\vec{\alpha} = d\vec{\omega}/dt$ is a vector. If the (angular) acceleration has only the effect of changing the magnitude of the angular velocity, then $\vec{\alpha}$ will be parallel (or anti-parallel) to $\vec{\omega}$, but if the orientation of the rotation axis also changes, then the direction of $\vec{\alpha}$ is given by the direction of the vector difference $\Delta\vec{\omega}$ . But note that rotations themselves cannot be defined as vectors, since they are not well behaved : when dealing with rotations, $A + B \neq B + A$

Relation between angular and linear quantities

NB : for the following expressions to be valid, angles MUST be measured in radians

Let's consider an arbitrary point on a rotating body, located at a distance r from the rotation axis. When the point covers an arc s, one has $\theta = s/r$ and, consequently:

$\omega = d\theta /dt = d(s/r)/dt = 1/r~ds/dt = v/r$

$\alpha = d\omega /dt = d(v/r)/dt = 1/r~dv/dt = a/r$

Using these results, one could readily obtain, for the case of a rotation with constant angular acceleration $\alpha$, kinematic expressions completely equivalent to the ones we had obtained for linear motion with constant acceleration:

$\theta (t)~=~ \theta_{0}+\omega_{0}t+1/2\alpha t^{2}$

$\theta (t)~=~ \theta_{0}+\omega_{av}t = \theta_{0} + 1/2(\omega_{0}+\omega_{final}) t$

$\omega(t)~=~\omega_{0}+\alpha t$

and, eliminating the time variable

$\omega^{2}(t)~=~\omega_{0}^{2}+2\alpha (\theta -\theta_{0})$

(see e.g. example 10.2 for a complete application of these formulae). Advice : even though rotational problems are formally identical to linear motion problems, conceptually they might appear less intuitive. It is recommended to do a few such problems to gain familiarity)

In the past, we had already learnt something about rotations, when we were dealing with circular motion. In fact, now we can define uniform circular motion as a motion characterized by constant angular velocity $\omega$. We had also learnt that uniform circular motion is characterized by a centripetal acceleration a_c=v²/r. Question: in a uniform circular motion with centripetal acceleration a_c, what is the angular acceleration $\alpha$ ????

The answer is clearly zero: uniform circular motion, by definition, has a constant $\omega$, and we have just seen that $\alpha$ represents the rate of change of $\omega$.... But not all circular motions are uniform, we can have a situation where the angular velocity is changing, which also means that the magnitude of tangential velocity for any given point of the rotating object will change, since $v_{t}~=~\omega r$. But if the magnitude of v_t is changing, we must have a tangential acceleration a_t, which, as seen before is related to the angular acceleration by $a_{t}=\alpha r$. Therefore, in general, the acceleration of a point on a rotating object is given by

$a~=~\sqrt{a_{t}^{2}+a_{c}^{2}}~=~\sqrt{(\alpha r)^{2}+(v^{2}/r)^{2}} = \sqrt{(\alpha r)^{2}+ \omega^{4}r^{2}} =$

$r\sqrt{\alpha^{2}+\omega^{4}}$

In summary:

• uniform circular motion has constant angular velocity, constant centripetal acceleration and zero tangential (and/or angular) acceleration
• rectilinear motion has zero centripetal acceleration, all acceleration is "tangential"
• non-uniform circular (or, in general, non-rectilinear) motion has a centripetal acceleration (with instantaneous value $\omega^{2}r$, where $\omega$ is the instantaneous angular velocity) and a tangential (therefore angular) acceleration. The global acceleration vector is given by the vectorial sum of centripetal and tangential accelerations.

Rotational Energy

How much kinetic energy is carried by a rotating object? To find this out, we first notice that, if an element of mass dm at a distance r from the axis of rotation is moving with velocity v, its kinetic energy is

$dK = \frac{1}{2}v^{2}~dm = \frac{1}{2}(\omega r)^{2}~dm$

The total energy of the rotating object will then be the sum of all the infinitesimal energies, i.e.

$K = \int dK = \frac{1}{2} \omega^{2}\int r^{2} dm = \frac{1}{2}I\omega^{2}$

where we define the moment of inertia $I = \int r^{2} dm$ . We conclude that, when dealing with a rotating object, we can express the (rotational) kinetic energy in terms of its angular velocity, provided we replace the mass with the moment of inertia. When dealing with a set of discrete objects rather than a continuous distribution, the moment of inertia is simply

$I = \Sigma_{i} m_{i}r_{i}^{2}$

where the r_i's are the distances of the masses m_i from the rotation axis. We can immediately draw some conclusions:

• when dealing with rotations the moment of inertia is somewhat the equivalent of the mass in linear motions. We can then expect that it will somehow represent the resistance a given object opposes to be put into rotation
• as it will become more and more clear, this "rotational inertia" will depend on the distance of the masses from the axis of rotation
• the moment of inertia is not an intrinsic property of a given object, since its numerical value depends upon the choice of axis of rotation

Calculation of I

How can one evaluate the integral $I = \int r^{2} dm$, since the differential is a mass while the function is a position? Obviously one needs to find a relation between masses and positions: remembering the definition of density $\rho = m/V =$ for an infinitesimal volume element = dm/dV, therefore

$I = \int r^{2} dm = \int_{V}\rho(r)r^{2}dV$

and, if the body is homogeneous,

$I = \rho\int_{V} r^{2} dV$

this might not always be easy to compute, but at least only contains spatial variables. When one or two dimensions of the object under exam are negligible (thin sheet or long thin rod) it is often useful to introduce the surface density $\sigma$ or the linear density $\lambda$

$\sigma$ = mass per unit area = $M/A~=~\rho t~~~$ t = thickness

$\lambda$ = mass per unit length = $M/L = \rho A$ L = length, A = area

Examples :

• thin rod of length L rotating around one end (similar to example 10.6) :
  
  if $\lambda = M/L$, then $dm = \lambda dL = \lambda dx$, therefore
  $I = \int r^{2} dm = \int_{0}^{L}\lambda x^{2}dx = \lambda L^{3}/3 = (M/L) L^{3}/3 = ML^{2}/3$
• thin rectangular metal plate (area = a b) rotating around side a
  
  $\sigma = M/A\rightarrow dm = \sigma dA = \sigma dx~dy$
  $I = \sigma \int_{A}r^{2}dx~dy = \sigma \int_{y=0}^{b}\int_{x=0}^{a} y^{2}dx~dy = \sigma \int_{0}^{a}b^{3}/3~dx =$
  
  $\sigma ab^{3}/3 = (M/A)A b^{2}/3 = Mb^{2}/3$

I will not expect you to be able to compute complicated integrals, but I do expect you to be aware and be able to apply the

Parallel Axis Theorem:

given the moment of inertia I_CM around an axis going through the center of mass, and the moment of inertia I around an axis going through an arbitrary point but parallel to it , the two are related by

I = I_CM+Md²

where d is the distance between the two axes

TORQUES

You have been told that an object is in equilibrium (i.e it has no acceleration) when the net force acting on it is zero. Obviously you were told a lie, since it is easy to think of many situations where a body is accelerating even if no net force is present (think e.g. of a water sprinkler with two jets spurting in opposite directions: the two jets exert equal and opposite forces, but when you turn the water on the sprinkler strats rotating...).

The reason for this ambiguity is that the statement about forces in equilibrium only applies to translations (or if you prefer, to point-like objects) and it ignores the possibility of rotations. How do then forces come into play when dealing with rotations, therefore angular, as opposed to linear, accelerations?

When a body is free to rotate around a point (or, to be more precise, around an axis), it is a well known fact that the force inducing rotations will be the more effective the larger is the distance between the point where the force is applied and the axis of rotation (compare what happens if you try swinging a door by pushing near the hinges or near the handle). Also, you will be much more effective in opening the door if the force is perpendicular rather than parallel to the door. We formalize these intuitive concepts introducing the following quantities:

• given a force, we call line of action the direction along which the force is acting
• given a force (with its line of action) and an axis of rotation, we call moment arm (or lever arm) the perpendicular distance between line of action and axis of rotation.
• finally we define as torque $\tau$ the product of force magnitude times lever arm, $\tau~=~Fl$.

By convention, torques are positive when they would cause a counter-clockwise rotation. Be clear that torques are not absolute quantities, but their numerical value depends upon the choice of axis of rotation.

The fact that when dealing with rotations what matters is not just the force but rather the torque has obviously been applied since antiquity: just think of wrenches, levers, etc...

If a torque is what causes rotations, then we can immediately guess that

a body will be in complete equilibrium (i.e. no accelerations) when both the net force and the net torque are zero

The first condition guarantees the absence of linear, i.e. translational, accelerations, while the second implies no angular, i.e. rotational, accelerations.

Newton's Law for Rotational Dynamics

Let us consider a body free to rotate around an axis, and let us apply to it a force F perpendicular to the axis and at a certain distance from it (so that the lever arm is not zero). For an infinitesimal volume of mass dm located at a distance r_i from the rotation axis we can write

dF = dm a

dF r = dm ar

$d\tau~=~dm~r^{2}\alpha$
$\tau = \int d\tau = \alpha\int r^{2} dm = I\alpha$

which is analogous to F = ma, provided one replaces forces with torques, masses with moment of inertia and linear acceleration with angular acceleration. Notice that an essential step in the derivation was the factoring of $\alpha$ outside the integral : while the linear acceleration is different for each element of the rotating body, the body as a whole has the same angular acceleration.

Among many other applications, the formula above allows us to estimate the moment of inertia for a body of arbitrarily complex shape. How ? by applying to it a known torque and measuring its angular acceleration.

Work, Power and all that

We have already seen the expression for kinetic energy associated with rotations:

$K_{rot} = \frac{1}{2}I\omega^{2}$

Let us recall what we had learnt in terms of work,power, etc. for translational motions :

• $W = \Delta K = \frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$
• $W = \int F_{x} dx$
• $\cal P$ = dW/dt = Fv

one could easily see that all of these expressions have their rotational equivalents :

• $W = \Delta K = \frac{1}{2}I(\omega_{2}^{2}-\omega_{1}^{2})$
• $W = \int \tau d\theta$
• $\cal P$ = $dW/dt = \tau\omega$