Uniform Circular Motion A special form of 2D motion is circular motion. Examples: ball on a string, cars exiting the highway, planets orbiting the sun Circular motion is accelerated motion, even if the speed of the object remains constant. WHY? The direction of the velocity changes at every point

Uniform Circular Motion • an object in uniform circular motion experiences a centripetal acceleration of magnitude: • centripetal acceleration always points toward the center of the circle.

Period and Frequency Uniform circular motion is periodic. Its behavior repeats after one period, T. The period, T, is the amount of time it takes for one complete revolution. The frequency, f, of motion is the rate with which the motion repeats. The frequency is the inverse of the period. per second seconds Hertz

Period and Speed The period and frequency can be related to the speed of an object in uniform circular motion. The object travels one circumference, 2πr, every period. Thus . . . . .

The Dynamics of Circular Motion v To remain in uniform circular motion, an object must be constantly accelerating toward the center of the circle. To constantly be center of the circle, the object must experience a net centripetal force. v

Horizontal circular motion ‘view from above’ Consider a box being whirled in a horizontal circle on a frictionless table. Draw the forces acting on the box. FT Fg and FN are not drawn since they point into and out of the paper respectively. (They are not important to the problem as they do not act into or opposite the direction of the acceleration) Fc = FT In what direction does this box accelerate? ALWAYS towards the center

Vertical circular motion ‘view from the side’ Consider a box being whirled in a vertical circle. Draw the forces acting on the box. Note: At the top of the circle, Fg would be added to FT Fg is now drawn since it is opposite to the direction the box accelerates. (There is no FN as it is not on a surface) FT Fg Fc = FT - Fg

A person is flying a 0.088 kg model airplane connected to a 10 m long string in a horizontal circle. The string exerts a force of 2.65 N on the plane. What is the plane’s speed? Given: ‘view from above’ m = 0.088 kg r = 10 m FT= 2.65 N v = ? m/s Fc = FT FT v2 2.65 = 0.088 10 v = 17.4 m/s

A pendulum 80 cm long is lifted above its equilibrium position and released. As the 0.050 kg pendulum bob passes through its lowest position the tension in the pendulum cord is 0.735 N, what is the pendulum's speed at this point? Given: ‘view from the side’ m = 0.050 kg Fc = FT - Fg r = 80 cm = 0.80 m FT = 0.735N v = ? m/s v2 0.735 – 0.49 = 0.050 0.80 FT ac v = 2.0 m/s Fg

A special case – just completing a vertical loop Unlike horizontal circular motion, in vertical circular motion the speed, as well as the direction of the object, is constantly changing. Gravity is constantly either speeding up the object as it falls, or slowing the object down as it rises. If we wanted to calculate the minimum or critical velocity needed for the block to just be able to pass through the top of the circle without the rope sagging then we would start by setting the tension in the rope to ____________. zero * ΣFc = Fg *only valid for the minimum velocity at the top to just complete a vertical loop.