# Uniform Circular Motion

## Presentation on theme: "Uniform Circular Motion"— Presentation transcript:

Uniform Circular Motion
The motion of an object at constant speed along a circular path. Centripetal Acceleration A vector quantity directed towards the center of curvature. Ac = v2/r When an object moves in a circle, its velocity is constantly changing but, it’s speed is not. Hence, the direction is always changing.

Centripetal Force A vector quantity directed toward the center of the circle. It magnitude is calculated by Newton’s second law. Fc=mac … ac = v2/r Fc=mv2/r Centripetal force is the net force acting on a body and is never in equilibrium.

Tangential Velocity As an object travels around the circle it’s velocity is drawn tangential to the curve. To find out the tangential velocity of an object going around the circle it is v = d/t where the distance is the circumference. vt = (2πr)/t

Centrifugal Force An object traveling in a circle behaves as if it is experiencing an outward force. This force, known as the centrifugal force, depends on the mass of the object, the speed of rotation, and the distance from the center. The more massive the object, the greater the force; the greater the speed of the object, the greater the force; and the greater the distance from the center, the greater the force. It is important to note that the centrifugal force does not actually exist. We feel it, because we are in a non-inertial coordinate system. Nevertheless, it appears quite real to the object being rotated. This is because the object believes that it is in a non-accelerating situation, when in fact it is not. For instance, a child on a merry-go-round is not experiencing any real force outward, but he/she must exert a force to keep from flying off the merry-go-round. Because the centrifugal force appears so real, it is often very useful to use as if it were real. The more massive the object, the greater the force. We know that this is true because an adult will have a harder time staying on a merry-go-round than a child will. The greater the speed of rotation, the greater the outward force. We know that this is true because a merry-go-round is harder to stay on, the faster it rotates. If you move further out on the merry-go-round, you will have to exert a greater force to stay on. In order to stay on a circular path, we must exert a force towards the center called centripetal (or "center-seeking") force. Consider a rope with a ball on the end. You can swirl the ball around in a circle over your head while holding onto the rope. The ball experiences the so-called centrifugal force, and it is the rope that provides the force to keep in moving in the circle.

Vertical Circles A ball is spinning in a vertical circle at the end of a string that is 2.0m long. If the ball has a mass of 3.5kg and moves at a constant speed of 8.0m/s find: The net force when the stopper is at the bottom of the circle. The net force when the stopper is at the top of the circle.

The net force when the stopper is at the bottom of the circle.
Fg = mg = (3.5kg)(9.81m/s2) = 34N Fc = mv2 / r = (3.5 kg) (8.0m/s)2 / 2.0m = 112N Fnet = Fc + Fg 112N + 34N 146N

The net force when the stopper is at the top of the circle.
Fg = mg = (3.5kg)(9.81m/s2) = 34N Fc = mv2 / r = (3.5 kg) (8.0m/s)2 / 2.0m = 112N Fnet = Fc - Fg 112N - 34N 78N

What is the minimum speed at which you could swing the ball so that the string won’t go slack (limp) at the top of the swing? Fg = Fc mg = mv2/r  "m" cancel out g = v2 / r v = √ (gr) = √ (9.81 x 2.0) v = 4.4 m/s