PHYSICS 12 Unit: Uniform Circular Motion
Imagine you have attached a rubber stopper to the end of a string and are whirling thestopper around your head in a horizontal circle. If both the speed of the stopperand the radius of its path remain constant, the stopper has uniform circular motion.Figure 1shows the position and velocity vectors at various positions during the motionof the stopper. Uniform circular motion also occurs if only part of the circle (an arc) iscovered.
Uniform circular motion occurs for the individual parts of any object spinning at a constantrate, for example, electric fans and motors, lawn mower blades, wheels (from thepoint of view of the centre of the wheel), and rotating rides at amusement parks. Circularor almost circular motion can also occur for objects or particles in orbits around otherobjects. For example, to analyze the motion of a planet orbiting the Sun, a satellitearound Earth, or an electron around a nucleus, we make the assumption that the motionis uniform circular motion even though the paths are often ellipses rather than circles.
As you learned in previous Chapter, an object travelling at a constant speed is undergoingacceleration if the direction of the velocity is changing. This is certainly true for uniform circular motion. The type of acceleration that occurs in uniform circular motionis called centripetal acceleration.Centripetal acceleration is aninstantaneous acceleration.
Recall that the defining equation for instantaneous acceleration is
To apply this equation to uniform circular motion, we draw vector diagrams and perform vector subtractions. Figure 2shows what happens to ∆v as ∆t decreases. As the time interval approaches zero, the direction of the change of velocity ∆v comes closer to pointing toward the centre of the circle. From the defining equation for instantaneous acceleration, you can see that the direction of the acceleration is the same as the direction of the change of velocity. We conclude that the direction of the centripetal acceleration is toward the centre of the circle. Notice that the centripetal acceleration and the instantaneous velocity are perpendicular to each other.
Figure 2
The Magnitude of Centripetal Acceleration
We can derive an equation for the magnitude of the centripetal acceleration in terms ofthe instantaneous speed and the radius of the circle. Figure 3shows a particle in uniformcircular motion as it moves from an initial position r1 to a subsequent position r2;its corresponding velocities are v1 and v2.
Example 1) A child on a merry-go-round is 4.4 m from the centre of the ride, travelling at a constantspeed of 1.8 m/s. Determine the magnitude of the child’s centripetal acceleration.
For objects undergoing uniform circular motion, often the speed is not known, butthe radius and the period (the time for one complete trip around the circle) are known.To determine the centripetal acceleration from this information, we know that the speedis constant and equals the distance travelled (2πr) divided by the period of revolution; T.
Example 2) Find the magnitude and direction of the centripetal acceleration of a piece of lettuce onthe inside of a rotating salad spinner. The spinner has a diameter of 19.4 cm and isrotating at 780 rpm (revolutions/minute). The rotation is clockwise as viewed fromabove. At the instant of inspection, the lettuce is moving eastward.
Example 3) Determine the frequency and period of rotation of an electric fan if a spot at the end ofone fan blade is 15 cm from the centre and has a centripetal acceleration of magnitude2.37 x 103 m/s2.
Understanding Concepts
1. How can a car moving at a constant speed be accelerating at the same time?
2. Calculate the magnitude of the centripetal acceleration a particle undergoing uniform circular motion at a speed of 4.0 m/s. 2.0 x 101 m/s2
3. At a distance of 25 km from the eye of a hurricane, the wind is moving at180 km/h in a circle. What is the magnitude of the centripetal acceleration, inmetres per second squared, of the particles that make up the wind? 0.10 m/s2
4. Calculate the magnitude of the centripetal acceleration in the following situations:
(a)An electron is moving around a nucleus with 2.1 8 x 106 m/s.The diameter of the electron’s orbit is 1.06 x 10–10 m. 8.97 x 1022 m/s2
(b)A cowhand is about to lasso a calf witha rope that is undergoing uniform circular motion. The time for one complete revolution of the rope is 1.2 s.The end of the rope is 4.3 m from the centre of the circle. 1.2 x 102 m/s2
(c)A coin is placed flat on a vinyl record, turning at 33rpm. The coin is 13 cmfrom the centre of the record. 1.6 m/s2
5. A ball on a string, moving in a horizontal circle of radius 2.0 m, undergoes acentripetal acceleration of magnitude 15 m/s2. What is the speed of the ball? 5.5 m/s
6. Mercury orbits the Sun in an approximately circular path, at an average distanceof 5.79 x 1010m, with a centripetal acceleration of magnitude 4.0 x 10–2 m/s2.What is its period of revolution around the Sun, in seconds? in “Earth” days? 7.6 x 106 s or 88 d