Forces and Motion in Two Dimensions Circular Motion

Objectives  Explain the acceleration of an object moving in a circle at constant speed  Describe how centripetal acceleration depends upon the object’s speed and the radius of the circle  Recognize the direction of the force that causes centripetal acceleration  Explain how the rate of circular motion is changed by exerting torque on it

Describing Circular Motion  The position of the object relative to the center of the circle is the position vector R.  During its path, the length of the vector doesn’t change, but the direction does. r1 r2 v2 v1

Describing Circular Motion  To find velocity, first find the displacement vector (∆r) or (r 2 -r 1 ) over a time interval (∆t.) ¯v = Δr / Δt  A moving object’s average velocity is ∆d / ∆t. But for an object in circular motion, the average velocity is ∆r / ∆t.

Describing Circular Motion  The velocity is at right angles to the position vector and tangent to the circular path.  ∆V is found by subtracting the vectors V 2 and V 1.

Centripetal Acceleration  Centripetal acceleration is the acceleration of an object in uniform circular motion and is always pointed toward the center of the circle.

Centripetal Acceleration  Acceleration is ∆V/ ∆t and is directed toward the center. a c = v 2 /r

Centripetal Acceleration  Period (t) = the time needed for an object to make a complete revolution, the circumference of the circle, 2πr.  The object’s speed is represented by: V = 2πr/t  So, substitute 2πr/t for V and the centripetal acceleration is: A c = (2πr/t) 2 / r or 4π 2 r / t 2

Torque  In relation to uniform circular motion, you have considered objects such as a person on a merry-go-round  These can be considered point masses  Now, consider rigid rotating objects  A mass that rotates around its own axis  Ex: doors

Torque  To open a door, you push at a distance from the hinges (axis of rotation) and in a direction perpendicular to the door  The product of the force applied to an object, the distance from the rotation point, and the direction of the force applied is called torque  Unit is for torque is Newton meters (Nm)

Torque  Solve for torque using the following equation: T = r FsinΘ T = torque r= radius F= force applied Θ= direction