Kinematics Uniform Circular Motion Objectives Define uniform circular

Kinematics Uniform Circular Motion

Objectives • Define uniform circular motion • Define period and frequency and compare them • State the directions of tangential velocity and centripetal acceleration. • Explain acceleration in g’s. • Solve problems for uniform circular motion

Uniform Circular Motion Circular motion at constant speed On Earth this normally involves horizontal circular motion. Point masses circling vertically speed up as they move downward, and they slow down as they move upward. As a result, they are not in uniform circular motion. Rigid objects, such as a rotating disk, can move in a vertical circles at constant speed.

Revolutions Linear motion is all based on linear meters. Meters traveled, meters /s of speed, and meters / s 2 of acceleration Circular motion, and other repetitive events, are based on the number of cycles the object moves through. There a number of ways to express cyclic behavior Revolutions, rotations, orbits, oscillations, vibrations, Etc. Of these revolutions is frequently encountered NOTE: Cycles, revolution, etc. do not technically have units. They are a count of an event. Unfortunately, reporting them as units varies. • In some cases revolutions appears as a variable in the equation, but revolutions is not mentioned when reporting the units. • However, there problems where revolutions are cited in the units.

Period The period, T , is defined as the time of one cycle. • Lower case t represents time that is NOT exactly one cycle. • Upper case T represents the time of EXACTLY one cycle. • If the time t of a number of revolutions (other than one revolution) is given, then the period can be solved by dividing the time by the number of revolutions. Usually the units of period are reported in seconds. Period is a measure of a very specific time, and seconds make sense. This is one of those cases where revolutions appears as a count in the equation, but is not reflected in the units.

Frequency The frequency, f , is the number of cycles in one second. • If the time t of a number of revolutions (other than one revolution) is given, then the frequency can be solved by dividing the number of revolutions by the time. The reporting of units for frequency varies. • It may be reported without citing revolutions: 1/s = s 1 = Hz (Hertz) • It could be reported as revolutions per second: rps (this is also Hertz) • And it is frequently reported in revolutions per minute: rpm In many cases the units rpm must be converted to rps = Hertz Divide rpm’s by 60

Period and Frequency Period and frequency are the inverse of one another

Example 1 Determine the period for a. Earth’s rotation 24 hours = 86400 seconds b. Earth’s revolution 365 days = 3. 15 107 seconds c. An object moving in a circle, and completing 20 revolutions in 60 s. d. An object with an angular speed of 20 rpm

Speed in Circular Motion Uniform circular motion involves constant speed. Why can’t we say constant velocity? The formula for constant speed is: Cyclic events are based on one cycle (one revolution) Distance of one cycle is the circumference: d = 2 r Time of one cycle is the period, T

Speed in Circular Motion Constant velocity does not exist in circular motion Constant velocity requires both Constant magnitude (constant speed) which you do have Constant direction: which you do not have However, at a specific instant of time objects do have an instantaneous velocity that points in a specific direction. v Instantaneous velocity is always directed TANGENT to a curved trajectory. As a result the velocity in circular motion is often referred to as tangential velocity. If an object every leaves the circular path, it will move in the direction of the tangential velocity.

Example 2 An object moving in a circular path of radius 5. 0 m completes 10 revolutions in 5. 0 s. Determine the objects tangential velocity.

Frequency Acceleration is said to be centripetal, ac Centripetal means center seeking (points toward the center) Deriving the following equation requires calculus and limits. As a result, the equation is simply given in this course. v ac Exams will often ask you to draw these two vectors at various points around the circular path. You may also be asked to explain the orientation of these vectors in words.

Example 3 An object moving along a circular path of radius 5. 0 m completes 10 revolutions in 5. 0 s. Determine the centripetal acceleration in m/s 2 and in g’s. This is a continuation of the previous problem. So far we have. Now solve for centripetal acceleration in m/s 2 Solving in g’s just involves a conversion: 1 g = 9. 8 m/s 2