Uniform Circular Motion Objects that Rotate Around an Axis

An object moving in a circle at constant speed is said to experience uniform circular motion. The magnitude of the velocity remains the same, but the direction of the velocity is always changing. As the object turns constantly, the velocity changes constantly. While acceleration is defined as change in velocity, with circular motion, the change in velocity does not involve change in magnitude but rather a change in direction.

The velocity changes constantly with acceleration being directed toward the center of the circle, centripetal acceleration.

To find the magnitude of the velocity of an object around the circle of radius r, The object will travel the circumference of the circle, 2πr each time it completes one revolution. That will be displacement, and the time to complete one rev is called the period. To convert rpm to period, rpm/60 sec = 1 rev/ x sec v = 2πr T A ball on a string 0.500 m long travels uniformly at a speed of 45 rpm, find the velocity of the ball. v = (6.28)(0.500)/1.33 = 2.36 m/s

To account for the change in direction of the object in the circular path, we look at how fast it is turning to remain in the circle. This acceleration is called centripetal or radial acceleration (directed along the radius). To calculate the centripetal acceleration, we can derive a formula and come up with ac = v2/r . Centripetal acceleration will always be perpendicular to v and point toward the center of the circle.

The moon orbits the earth at a radius of 384,000Km with a period of 27 The moon orbits the earth at a radius of 384,000Km with a period of 27.3 days, Find the velocity of the moon and the acceleration of the moon towards the earth. v = 6.28 (3.84 x 108m) (27.3)(24)(3600) v = 1020 m/s a = v2/r = 1020 2 /3.84 x 108m a = 2.72 x 10-3 m/s2 A 150 g ball at the end of 0.600 m cord revolves uniformly in a horizontal circle making 2 revolutions every second. Find the centripetal acceleration of the ball. V = 2πr/T = 6.28)(0.600)/0.500 s = 7.54m/s ac = v2/r = (7.54)2/ 0.600 = 94.8 m/s2

Centripetal force is often misinterpreted as a force pushing outward on the circling object, or on an object in a car going around a curve. The fictitious centrifugal force is the perceived result of you continuing in a straight line while the car curves. Upon release, will the ball fly straight out or tangentially?

The frictional force on the car is what causes the centripetal acceleration. To the right would be a front view where the car would be turning to our right. This concept is involved when engineers design roads and the sharpness of curves in the roads.

A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 14 m/s. Will the car make the turn, or skid if a) the pavement is dry and μ = 0.60, or b) the pavement is icy and μ = 0.25? a) Fn = 9800 N so Ffr = 0.6)(9800) = 5880 N so long as Fc < 5880 N, the car will not skid Fc = m ac (Newton’s 2nd law) = 1000 kg) (14m/s)2 = 3920 N 50 m

In case b) the frictional force is (9800)(0 In case b) the frictional force is (9800)(0.25) or 2450 N since the Fc must be 3900 N, the car will skid and fly off the curve in basically a line. What is the maximum speed that the car can travel and make it around the curve in each case above. a) 5880 N = 1000 v2/50 m; v = 17.1 m/s b) 2450 N = 1000 v2/50 m; v = 11.0 m/s

In order to allow for faster speeds on curves, engineers plan an build banked curves. When planning, the engineers design curves so that no friction is required to make the turn. The bank will allow for smaller centripetal forces and thus faster speeds.

To determine the angle for a banked turn of a particular radius and speed of the vehicle, we know that the Fc = Fn sin Θ Since there is no vertical motion Fn = mg/cos Θ so that (mg/cos Θ)(sin Θ) = m v2/r mg tan Θ = m v2/r g tan Θ = v2/r Θ = arctan v2/rg So for a freeway off-ramp of radius 50 m and a design speed of 14 m/s, Θ = 22 °.

Vertical circles When a mass is swinging in a circle at the end of a string, the tension in the cord provides the force to cause ac. The FBD shows the forces when the circle is vertical. The main points of concern will be the minimum speed required to complete the circle at the top and the tension in the cord at the bottom.

A ball of mass 150 g on the end of a 1 A ball of mass 150 g on the end of a 1.10 m cord is swung in a vertical circle. Determine the minimum speed required by the ball at the top of the circle so that it will not fall, and the tension at the bottom of the circle when it is moving at twice the speed at the top. Σ F = ma Ta + mg = mac = m (v2/r) If T = 0 the ball will fall when at the top, so, mg = m (v2/r) and (gr)1/2 = V = [(9.80)(1.10)]1/2 = 3.28 m/s

At the bottom, V = 6.56 m/s, and ΣF = m (v2A/r) TB – mg = m (v2A/r) And T = m (v2A/r) + mg = 7.34 N