Chapter 10: Rotational Motion

Adria Updike

10 Chapter 10: Rotational Motion

Textbook Section 4.4: Uniform Circular Motion

Textbook Section 10.1: Rotational Variables

Uniform Circular Motion refers to motion in a circular path at a constant speed. Examples of this include carousels, racetracks, and stellar orbits. Rotational motion refers to objects moving in a circle around a central point, and translational motion refers to motion in a straight line. You can combine these types of motion; for example, putting a spin on a football as you throw it.

The circumference of a circle is the distance around the edge, and is calculated as $C = 2 \pi r$ where $r$ is the radius of the circle. The angle through which a clock hand might move is given at $\theta = s/r$ , where $s$ is some arc length (distance swept out by the tip of the clock hand in some given time interval).

A radian is defined at 360 $^{\circ}$ of a circle divided by $2\pi$ , or 57.3 $^{\circ}$ . $2\pi$ radians are equivalent to one revolution (one full trip around the circle and back to your starting place).

The angular velocity of an object moving in a circular path tells you how many $\theta$ you move through in a certain time period, and is calculated as:

$\omega_{av,z} = \frac{\Delta \theta}{\Delta t}.$

This is the average angular velocity of an object spinning in the $x-y$ plane (the direction of the angular velocity is perpendicular to the plane, or in the direction of $z$ ). Instantaneous angular velocity $\omega_z$ is found as following:

$\omega_z = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d \theta}{d t}$

The units of angular velocity are given in radians per second (rads/s). The angular displacement of an object moving in a circular path is given by $\Delta \theta = \theta_f - \theta_i$ .
Note that $\Delta \theta$ can be a number larger than 2 $\pi$ ! Typically when moving in a circle, we start over when we reach our starting point again, but not when calculating angular displacements.

Clockwise motion (cw) around a circle gives us a negative angular velocity, and counterclockwise motion (ccw) gives us a positive angular velocity. We assume for now that our rotating objects are rigid bodies (do not deform, bend, or stretch while moving in a circle) and that every point on the object has the same angular velocity.

Angular distance $\theta$ can be given at $\theta = s/r$ , and velocity is equivalent to $v = d/t$ . We can re-write the former equation as $s = \theta r$ , and plug into the latter to receive $v = \theta r/t = \omega r$ . Now we can relate translational (also called linear) velocity $v$ and our angular velocity $\omega$ if we know the radius of the circle traced out by our object.

Textbook Section 10.4: Moment of Inertia and Rotational Kinetic Energy

Our moment of inertia $I$ was defined in Chapter 9 in terms of the torque, $\tau$ . We found that $\tau = I \alpha$ , and we could write $I$ for a collection of point particles as

$I = \sum_i m_i r_i^2$

where $m_i$ is the mass of a point particle and $r_i$ is the distance between the point particle and the axis of rotation (pivot point). Calculating $I$ for more extended objects than point particles require us being careful as to which axis we are considering. All lines drawn through a point particle are identical, but you can draw different lines through a cylinder and rotate it around different axes of rotation, changing its particle distribution from the axis and thus its moment of inertia.

You can find the moment of inertia for various object configurations in your textbook. A few are listed on the next page. To construct an object made up of many discrete pieces, each of which has its own moment of inertia, you merely add the moments of inertia of all the pieces. You will never have a negative moment of inertia.

What if you want to use an axis not given in your table? Take a look at the solid sphere in the left bottom corner of the diagram on the previous page. What if you wanted to put the axis of rotation not through the center of the sphere, but through an edge of the sphere? That’s when you need the parallel axis theorem. The parallel axis theorem tells you how to modify your given, center of mass moment of inertia for an object in order to rotate the same object through a different axis.

$I_{pp} = I_{cm} + M d^2$

You begin the parallel axis theorem using the given center of mass moment of inertia from the table for that shape, and simply add on another term that involves the mass of the object times the square of the distance between the new axis and the axis used in the center of mass moment of inertia. For the example given, our new moment of inertia for our sphere rotating around an edge instead of the center would be $I_{pp} = 2/5 \,\, M R^2 + M R^2$ because the mass of the sphere is $M$ and the distance between the axis used in the picture and the edge of the sphere is another factor of the radius $R$ .