1. Rotational Motion

2. ---- = ------------ Uniform Circular Motion r r v2v2 v1v1 7v7v 2 2 a b Chord ab c d e f Using similar triangles abc def Where r is proportional to v 1 or v 2 7v7v ---- = ------------- v Chord ab r Chord ab = d = v t 7v7v v v t r Since a = ------ 7v7v t 7v7v t v x v r ---- = ----------- acac = ----- v 2 r a c = Centripetal Acceleration The direction of the centripetal acceleration is always toward the center of the circle

3. Difficult to directly measure velocity of an object moving in a circle. Useful to measure the “time for one complete revolution”. Defined as the Period (T). The distance traveled in a circle is the circumference 2 r of the circle.

4. Can now calculate the acceleration of an object moving in a circle knowing the Period and radius of the circle. Centripetal Acceleration )(

5. Centripetal Force If an object is moving in a circular path, there must be a change in its velocity and hence an acceleration. If there is an acceleration, a force must be present to cause this acceleration. It also must be in the same direction.

6. Circular Motion Equations Period = The time for one revolution Sometimes useful to show the number of revolutions per time Known as frequency Frequency is the inverse of the Period Units are

7. Problem : A 75.0 kg person is attached to a pole by a 5.00 meter rope. He makes one revolution in 4.50 seconds. Find : The speed of the person His Acceleration The Force required to keep him on his path

8. Problem 2: A car approaches a level, circular curve with a radius of 45.0 m. If the concrete pavement is dry, with a coefficient of friction of 1.20 between the rubber and concrete, what is the maximum speed at which the car can negotiate the curve at a constant speed? The car will be in uniform circular motion on the curve, so there must be a centripetal force. This force is supplied by friction, so the maximum frictional force provides the centripetal (net) force when the car is at its maximum tangential speed.