Rotational Motion

We have now covered the three basic elements of mechanics; force, energy and momentum. In all of our discussion so far we have considered an object moving from one location to another. However, there are other types of motion than this. For example and object can rotate, the object is clearly moving, but the center of the object stays at the same location. The motion we have discussed up to now is called translational motion, when the center of mass of the object moves location. We now will discuss rotational motion, when an object rotates around a fixed point. Of course objects can have both translational and rotational motion and the same time and often do. We separate them here just to make it easier to focus on the rotational part.

All of the physics we have learned still applies in exactly the same way for rotational motion.  The major difference in rotational motion is that the cartesian coordinate system we have been using of x,y,z doesn’t make as much sense. Instead it makes more sense to use polar coordinates R,Θ.  If an object is in fixed rotation then its distance to the axis of rotation, R, does not change. The only coordinate that changes with time is the angle, Θ. To describe rotational kinematics we would therefore want to use an angular velocity and angular acceleration defined by:

\vec{\omega} = {d\vec{\theta}\over{dt}}              \vec{\alpha} = {d\vec{\omega}\over{dt}}

For angular quantities we will want to use radians instead of degrees, so a full circle is 2π instead of 360º.  The units of Θ will be radians or rad, for angular velocity, ω, they will be rad/s and for angular acceleration, α, it will be rad/s².  If we consider one point on our rotating object then we can connect angular quantities to translational quantities by multiplying by R for that point.  So the distance the point travels would be RΘ, the velocity of that point would be Rω and the tangential acceleration of that point would be Rα.  Note that the point already has a centripetal acceleration because it is undergoing circular motion.  The tangential acceleration is the the change in velocity orthogonal to the centripetal acceleration.

The above equations for angular velocity and acceleration are identical to equations we had in translational kinematics with just changes in the variables.  This means solutions to these equations must also be identical.  So for constant angular acceleration we have:

\omega = \omega_0 + \alpha t                                 \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2
\omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0)                              \bar{\omega} = \frac{1}{2}(\omega_0 + \omega)

Now that we have discussed angular kinematics let us turn to angular force which is called torque.  Torque must be defined so that it is proportional to the angular acceleration.  To do this we must define torque as:

\vec{\tau} = \vec{R} \times \vec{F} = RF\sin{\phi}

where torque is the cross product between the a radial vector from the axis of rotation and the force.  Since we haven’t discussed cross products yet, for now we can just define torque as the magnitude of the distance to the axis of rotation times the magnitude of the force times the sin of the angle between them. Notice we have defined that angle as φ because we are using Θ as a coordinate.  Now that we have torque and angular acceleration we only need the angular equivalent of mass to get Newton’s second law.  This is called the moment of inertia of an object and is given by:

I = \sum_i m_i R_i or I = \int \rho(R)R^2 dR

For all simple geometries this integral was solved long ago and you can just look up what common moments of inertia are. For example for a disk it is I=½MR².  You can find more than you will need on Wikipedias List of moments of inertia.  With torque and moment of inertia defined we can now right Newton’s second law for rotating objects.

\sum \vec{\tau} = I \vec{\omega}

Here is an example of torque from the movie The Dark Knight.

If an object is rotating then it will take work to stop that rotation. That means there must be rotational kinetic energy.

KE = \frac{1}{2} I \omega^2

Which is just what you would expect replacing the translation quantities by their rotational equivalents.  Rotational kinetic energy is just another form of energy, when energy is conserved the rotational kinetic energy must come from or go to potential energy. Or rotational kinetic energy can be converted into translational kinetic energy.  Often an object will have both translational and rotational motion, like the wheel of a bike.  There is a special case of this called rolling without slipping.  This occurs when the objects rolling causes it to move. So then the distance it goes x = RΘ, it has velocity v=Rω and acceleration a=Rα.  Note that this is different then when we had the same formulas above; here we are talking about the center of mass motion of the object. Before we were just talking about the motion of a point on a rotating object whose center is at rest.  Here is an example of rotational kinetic energy and rolling without slipping from the movie the Princess Bride.

Finally we get to angular momentum.  Angular momentum is related to the net torque just like momentum was related to net force.  The letter L is commonly used for angular momentum. The angular momentum of a rotating object is given by

\vec{L} = I\vec{\omega}.

When we introduced translational momentum we saw that Newton’s second law could be written in terms of the derivative of momentum. The same thing now holds for angular momentum \sum \tau = dL/dt.  And thus if there is no net torque on an object the angular momentum is conserved. Note that there could be a net force on an object, but no net torque or a net torque but no net force. So angular momentum conservation doesn’t imply momentum conservation or vice versa. If we have whose center of mass is moving we can still discuss its angular momentum. In this case is depends on what axis of rotation one is considering and then the angular momentum is given by

\vec{L} = \vec{R} \times \vec{p} = Rmv\sin{\phi}