Lecture 11

Rotational Kinematics

You might not have realized it, but until now all the objects whose motion we studied (cheetahs, rabbits, airplanes, sailboats, 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. Mathematicians like the radian because it has a non-arbitrary definition, as opposed to ,e.g., the degree, which is defined by the arbitrary choice of dividing the circumference into 360 equal parts (and one might ask why not 100, or 60 or any other partition). On the other side, 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 quantities.

Useful to remember :

• angle of a full circle : = $2\pi r/r = 2\pi$
• angle of a half circle = $\pi$
• right angle = $\pi /2$
• being both linear measurements of angles, radians and degrees must be proportional to each other, i.e. $\theta$(radians)/$\theta$(degrees)=constant. From this we can get:
  1 radian / ?? degrees = 2$\pi /360^{0}$, i.e.
  
  1 radian = $360^{0}/2\pi = 57.3^0$

Rotational Variables

In the past, we have learnt how to analyze translational motions in terms of displacements, velocities (both average and instantaneous) and accelerations (and we have typically limited ourselves to the case of constant acceleration). Rotational motion is handled in a completely analogous way, provided we introduce the appropriate quantities to describe rotations, i.e. :

• angular displacement : 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...). It is wise to get used to measuring angular displacements in radians rather than degrees
• angular velocity $\omega : \Delta\theta / \Delta t$, i.e. ratio between angular displacement and time to achieve that displacement. According to the context, $\omega$can be an average or an instantaneous velocity. 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?
  
  Using the "trick" we have learnt in the first class we can write:
  rev/sec = rev/sec x rad/rad = rad/sec x rev/rad = $\omega_{rad/sec}\times 1/(2\pi )$
  If we also recall our definition of period T, we also have that
  $\omega_{rev/sec} = 1/T$
• angular acceleration $\alpha: \Delta\omega /\Delta t$. If the angular velocity of a rotating object is not constant, then we are in the presence of an angular acceleration, with obvious definition

In a manner completely analogous to what we had done earlier for translational motions, we could derive all the useful expressions to describe rotations. Assuming constant (or at least average) angular acceleration $\alpha$, we have:

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

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

Also, eliminating the time variable

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

More quantities related to rotations

When a merry-go-round turns at a constant rate, each one of its points moves with the same angular velocity $\omega$. But are all points moving equally fast? Certainly not ! Suppose the platform makes a full turn in T seconds, we have seen that T = 2$\pi/\omega$. Now take a point at some distance r from the center. In T seconds, the point covers a distance $2\pi r$. By definition its speed is $v=2\pi r/T~=~ \omega r$. Points at different radii are therefore moving with ("linear") velocity proportional to the radius, i.e points near the periphery are moving much faster than points near the center. The direction of the velocity vector is, at any instant, along the tangent to the circle, we will then refer to it as tangential velocity. In summary

$v_{tangential}~=~2\pi r/T~=~\omega r$

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. But "v" is just what we have called the tangential velocity, therefore we immediately get

$a_{c}~=~\omega^{2}r$

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 $v_{tangential}~=~\omega r$. But if the magnitude of v_tangential is changing, we must have an acceleration, which we can call a_tangential (do you remember what I had told you in Lecture 7, i.e. that an acceleration perpendicular to the velocity vector changes the velocity's direction, while a parallel acceleration changes its magnitude ?). We can immediately relate the tangential acceleration to the angular one :

$a_{t}~=~(v_{t}-v_{0})/t ~=~(\omega r-\omega_{0}r)/t = r\alpha$

A useful mnemonic rule : linear quantity = angular quantity x radius :

$s=\theta r,~v=\omega r,~a=\alpha r$

which are valid provided angular displacements are measured in radians. 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.

Re-discovering the wheel

A vehicle moving on wheels, is a common example of a translational motion being due to a rotation. What is the relation between the translational velocity of the wheeled vehicle and the angular velocity of the wheels (assuming you are not "spinning your wheels", i.e. the wheels are not slipping on the surface)?

You could easily verify (e.g. by winding a string around a wheel and letting it unroll as the wheel turns) that when the wheel makes a full turn the vehicle advances by an amount equal to the wheel circumference. We then have, if the suffix v refers to the vehicle and w to the wheel,

v_v=s_v/t=(in one turn)$=2\pi r/t = \theta_{w} r/t = \omega_{w}r$

Another way to state the same result would be to say that the translational speed of the vehicle is equal to the tangential speed of the wheel's rim (as measured with respect to the axle, not with respect to the ground).

With similar arguments, one could also see that $a_{v}=\alpha_{w}r$