Dynamics of Uniform Circular Motion

Dynamics of Uniform Circular Motion • Uniform Circular Motion is the motion of an object traveling at a constant (uniform) speed on a circular path. The period (T) is the time it takes an object to make complete circle or revolution. Because velocity = distance / time the velocity of an object around a circle is v = 2πr (the distance around a circle) T (time it takes for one revolution) This fan has a radius of .7m and is rotating at 90 rev/min. How fast is the end tip of the fan blade going?

This fan has a radius of .7m and is rotating at 90 rev/min. How fast is the end tip of the fan blade going? Convert 90 rev / min to the time it takes for one revolution. V = 2πr = 2(3.14)(.7m) = m/s T 1.67 sec r = .7m

Centripetal Acceleration • Uniform circular motion emphasizes that velocity vector is constant – but the direction of this constant velocity is not constant. • By definition, acceleration is the change in velocity over time. The “change in velocity” for an object in UCM is the change in the velocity vector. • The acceleration that occurs is toward the center of the circle. It is the force that is pulling the object “in”

Centripetal Acceleration • The centripetal acceleration (the acceleration toward the center of a circle) of an object depends on the velocity of the object and the radius of the circle and is expressed as: ac = v2 r d = 100m How fast would the outer most part of the space station have to spin in order to have a centripetal acceleration of 9.8 m/s2

How fast would the outer most part of the space station have to spin in order to have a centripetal acceleration of 9.8 m/s2 ? d = 100m a = v2 / r a = 9.8 m/s r = 100m/2 = 50m Solving for v: v = \/~(a r) = \/~(9.8m/s2)(50m) = 22.1 m/s A bobsled goes down a track, moving at a constant velocity throughout all three turns. At which turn will there be the greatest centripetal acceleration? A B C D – All the same If the bobsled moves at 35 m/s throughout the course and turn B has a radius of 20m and turn C is 50m how many G’s of acceleration is the sledder feeling through turns B & C ?

Centripetal Force • The net force required to keep an object of mass m, moving at a constant velocity v, on a circular path. D B F = mv2 r Note that v2/r = acceleration A C At what point will a person feel the most centripetal force? A B C D E= all the same A 50 kg person and a 100 kg person ride the “Enterprise” at Valleyfair. The ride has a radius of 7 meters and takes 4 seconds per revolution. How much centripetal force does each person feel at point A & C? “Gondolas start out hanging from the ride, then go to a horizontal position by centrifugal force from the spinning ride, which then lifts to a vertical position turning riders upside down.”

Centripetal Force Lab Review A person on a centrifuge with a radius of 6m spins around once in 2 seconds. The person is in contact with a scale that is reading 1,100 lbs. What is the mass of the person being spun? How many G’s is this person experiencing? 1N = .22 lb 2.2lbs = 1kg

A person on a centrifuge with a radius of 6m spins around once in 2 seconds. The person is in contact with a scale that is reading 1,100 lbs. What is the mass of the person being spun? How many G’s is this person experiencing? Start by finding the velocity of the centrifuge: V = 2πr / T T=2sec r= 6m v= 2(3.14)(6) / 2 = 18.8m/s Then, using Fc = mv2/r, rearrange the equation solving for m m = r Fc / v2 convert 1,100 lbs to Newtons 1,100 lbs x 1 N = 5000N 1 .22 lbs m = 6 m(5000 N) / 18,82 = 84.9 kg = mass ac = v2 / r = 18.82 / 6 = 58.9 m/s2 58.9m/s2 / 9.8m/s2 = 6.0 G’s of acceleration

Is it harder for the man to hold his partner when the partner is hanging straight down and is stationary or when the partner is swinging through the straight down position? It is harder for the man to hold his partner when the partner is swinging through. When stationary, the man only needs to hold the force of the partner against gravity . Ex. 50kg x 9.8m/s2 = 490 N When the partner swing, the man needs to hold with the force from gravity, PLUS the centripetal force of the partner because they are accelerating (changing v)

When a car turns a corner, there is a centripetal force toward the center of the circle. The force that is being applied to the center of the circle is coming from static friction between the car and the ground. If the rubber on the car cannot handle the centripetal force, the car will slide. A soapbox racer is making a turn on asphalt at 20 m/s. The coefficient of frictionstatic is .9. The radius of the turn is 40 m. Will the soapbox crash into the hay? Why or why not? Static Friction is keeping the car from sliding Fc = μFN = mv2/r μ mg = mv2/r V2 = μmgr / m = μgr V = sqrt .9(9.8)(40) = 18.8 m/s max speed around the turn

Banked Turns When a car turns a flat corner, static friction provides the centripetal force. The need of static friction can be eliminated by making a bank in the turn. The centripetal force becomes a vector component of the normal force. In the case above, it is = FN(cos Ɵ). If there is a set amount of bank, the can be a perfect speed to take the turn at without needing static friction. But, if you go too slow, you may slip, too fast and you may skid

Banked Turns Tan Ɵ = v 2 r g *this equation indicates that for a given speed, the centripetal force needed for a turn of radius r can be obtained by banking the turn at an angle, independent of the mass of the vehicle. Mr. Shaf hot A racetrack is designed with a 40o bank angle and a radius of 240 m at the ends. At what speed is the track designed for cars to make an ideal turn? V = sqrt (tan Ɵ)rg = sqrt (tan 40)(240m)(9.8m/s2) = 44.4 m/s

A racecar enters a banked turn that has a 40o angle. The car is going 34 m/s into the turn. At what distance should the driver locate his car if he wishes to take the turn without depending on friction? 140.5m d = 140.5 sin 50 = 183.4 m d mg 50o 40o Using the equation tan Ɵ = v2 / rg we want to find the radius of the turn. Once we have the radius, we’ll us trig to find d r = v2 / tan Ɵ(g) = 342 / tan 40 (9.8) = 140.5 m

FN sin 50 – mg = 0 (no vertical acceleration) A racecar enters a banked turn that has a 40o angle. The car is going 34 m/s into the turn. At what distance should the driver locate his car if he wishes to take the turn without depending on friction? 50o FN Fc =FN(Cos Ɵ) = mv2 / r 50o 140.5m d = 140.5 sin 50 = 183.4 m d mg 40o Fc = FN cos Ɵ = mv2 / rif we can find the radius of theturn, we can find d using trig. Solving for r, r = mv2 since FN = mg/sin50, we’ll substitute this in FN cos Ɵ r = mv2 = v2 (sin50) = 342 (sin50) = 885 = 140.5m mgcosƟ g cosƟ 9.8(cos 500) 6.3 (sin50)

Satellites in Orbit There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius. The centripetal force acting on satellites is gravity: Fc = G m1m2 = mv2 r2 r v = GME T = 2πr3/2 r GME

A rocket is launched to a height of 50 miles above the earth’s surface. How fast does the rocket need to be going in order to stay in orbit? (1 mile = 1.6 km) (Earth’s radius = 6,380,000 m) (Earth’s mass = 5.98 x 1024) V = GME r Synchronous satellites are satellites that orbit the earth with no movement relative to the earth’s surface. How many miles per hour is the earth orbiting the sun? (93,000,000 miles radius) (mass of sun = 2 x 1030) V = GME r

Artificial Gravity Artificial gravity can be created by centripetal force. If this spaceship has a radius of 3000 m, how fast does the outside of the spaceship have to spin in order to simulate a gravitational acceleration of 9.8m/s2? A = v2 / r

Vertical Circular Motion Vertical circular motion occurs when an object moves in a vertical path. The speed of the object often changes from moment to moment as a result of gravity. If an object is in freefall, the centripetal force is gravitational force (mg). mg If an airplane does a parabolic downward turn to simulate artificial weightlessness, what would the radius of the turn have to be if the airplane flew the turn with a speed of 100 m/s (220 mi/hr)? mg

Dynamics of Uniform Circular Motion