Circular Motion and Simple Machines

The three great technological advances of human civilization are said to be the discovery of fire, the invention of the wheel, and the invention of sliced bread. A discussion of the importance of sliced bread belongs in another class (perhaps a baking class). Fire will be discussed next week. This brings us to the wheel. In this section, we are going to consider two types of circular motion. The first type is when an object moves in a circle around an external axis. The second type is that of an extended object that is rotating about an axis that goes through the object. Both types of circular motion can be analyzed with the same mathematical tools if angles are used as the variables instead of lengths. First, we should review some trigonometry.

Tutorial
Click the Circular Functions animation to learn about the relationship between angles, trigonometric functions, and motions around a circle.

An angle is a measure of the rotation between two radii of a circle. One revolution (rev) is one complete rotation. One rev is 360°, one half of a rotation is 180°, and so on. A radian is the measure of an angle that relates the length of the arc to the length of the radius, so in radians, the angle is the following.

$\theta =\frac{s}{r}$

There are 2p radians in one rev, and one radian is approximately 57.3°. Warning: Many of the equations that we will use in this section only work if the angles are in radians. Be sure to convert angles to radians first.

Example: What is an angle of 60° in radians?

$\begin{array}{l}\theta =60°\\ \theta =60°\times \left(\frac{\pi \text{ rad}}{180°}\right)=\frac{\pi }{3}\text{ rad}\end{array}$

The angular velocity w is defined as the change in angle over the change in time.

$\omega =\frac{{\theta }_f-{\theta }_i}{t_f-t_i}=\frac{\Delta \theta }{\Delta t}$

The units for the equation above are newton-metres, a unit which is the same as a joule. If there is no displacement, the work is 0. If the force is perpendicular to the displacement, the work is 0. If the force points in the same direction as the displacement, then the work is positive and the energy of the system increases. If the force points in the opposite direction from the displacement, then the work is negative and the energy

$$\Delta E=E_{final}-E_{initial}=W_{total}$$

If the object has a constant angular velocity, then the motion is called uniform circular motion. The magnitude of the velocity tangent to the circle can be calculated by the following.

$v=\frac{2\pi r}{T}=\omega r$

In this equation, r is the radius of the circle, and T is the time required for the object to complete one revolution.

Example: A wheel with radius 0.250 m is spinning at 275 rpm. What is the angular velocity in radians per second and the linear speed of a point on the rim?

$\begin{array}{l}\omega =\left(275\frac{\text{rev}}{\text{min}}\right)\left(\frac{1\text{ min}}{60\text{ sec}}\right)\left(\frac{2\pi \text{ rad}}{\text{1 rev}}\right)=28.8\text{ rad/s}\\ v=\omega r=\left(28.8\text{ rad/s}\right)\left(0.250\text{ m}\right)=7.20\text{ m/s}\end{array}$

Even when an object is in uniform circular motion, the velocity is changing because the direction is constantly changing. If the velocity is changing, there must be an acceleration causing the change. This acceleration is called the centripetal acceleration because it always points in toward the center of the circle. If there is acceleration, then there must be an unbalanced force that points in the same direction as the acceleration. This unbalanced force is the cause of the circular motion and is called the centripetal force. The magnitude of the centripetal acceleration and the centripetal force can be calculated from these formulas.

$\begin{array}{l}{a}_{cen}=\frac{v^2}{r}\\ F_{cen}=\frac{m{v}^2}{r}\end{array}$

You can feel this for yourself. Securely tie a small ball or other soft object to a length of string. Go outdoors and carefully swing the ball in a circle over your head. Notice that in order to keep the ball moving in a circle at a constant speed, you have to pull the string in toward the center of the circle. You are supplying the force needed to keep the ball moving in a circle. Be careful. The equations above do not indicate the causes of the circular motion but merely give the relationship between the force and acceleration and the linear speed of the rotating object.

Example: An automobile with a mass of 1,640 kg is driving around a curve with a radius of 25.0 m at a speed of 15.0 m/s. What is the magnitude of the centripetal force on the automobile? What is the cause of the centripetal force on the automobile?

$F_{cen}=\frac{m{v}^2}{r}=\frac{\left(1640\text{ kg}\right){\left(15.0\text{ m/s}\right)}^2}{\left(25.0\text{ m}\right)}=14800\text{ N}$

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There always is a centripetal acceleration and centripetal force whenever an object is moving in a circle. If there is a change in the angular velocity w, then there is an additional acceleration and force tangent to the circle. If w is changing, then we can define the angular acceleration a by the following.

$\alpha =\frac{\Delta \omega }{\Delta t}$

If the angular acceleration is a constant, then the equations for angular motion look very familiar.

$\begin{array}{l}\theta ={\theta }_i+{\omega }_i\Delta t+\frac{1}{2}\alpha {\left(\Delta t\right)}^2\\ \omega ={\omega }_i+\alpha \Delta t\\ 2\alpha \left(\Delta \theta \right)={\omega }_f{}^2-{\omega }_i{}^2\end{array}$

These equations have exactly the same form as the equations we used for linear motion in Week 2. They are used in the same way, as well.

Example: A rotating pulley 24.0 cm in diameter has an initial angular velocity of 30.0 rad/s. The angular velocity steadily increases to 42.0 rad/s over a period of 6.00 seconds. What is the angular acceleration of the pulley?

$\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}=\frac{\left(42.0-30.0\text{ rad/s}\right)}{6.00\text{ s}}=2.00\frac{\text{rad}}{{\text{s}}^2}$

We learned back in Week 3 that unbalanced forces are the cause of accelerations and that mass is the resistance an object offers to the force. In exactly the same way, unbalanced torques & tau; cause angular accelerations, and the moment of inertia I is the resistance to the torque. A torque is the product of the force used to cause the rotation and the moment arm st. The moment arm is the perpendicular distance from the axis of rotation to the point of application of the force.

The magnitude of the torque can be calculated by $\tau =F{s}_t$ , and the relationship between torque, angular acceleration, and moment of inertia is $\tau =I\alpha$ . This equation is the rotational version of Newton's Second Law and applies to objects rotating about a fixed axis.

If a torque is applied on an object that rotates through an angle, then work is done by the torque. The power is then the amount of work done over a period of time.

$\begin{array}{l}\Delta W=\tau \Delta \theta \\ P=\frac{\Delta W}{\Delta t}=\tau \frac{\Delta \theta }{\Delta t}=\tau \omega \end{array}$

The power applied by a torque then can be calculated by the product of the torque and the angular velocity.

Example: Find the angular velocity of a motor that applies 130.0 Nm of torque at a power of 650.0 W.

$\omega =\frac{P}{\tau }=\frac{\left(650.\text{ W}\right)}{\left(130.\text{ Nm}\right)}=5.00\text{ rad/s}$