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09/21/09 - Circular Motion
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 | Newton's thought experiment: Launching a projectile horizontally from a mountain top
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 | If a projectile is launched horizontally from a mountain top, it will follow a trajectory that eventually touches the earth. As the initial launch speed is increased, the projectile will hit the ground farther and farther from the mountain top.
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| If the launch speed is large enough (thousands of meters per second), then we have to take the curvature of the earth into account.
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| We find that when the launch speed is about 7920 m/s (about 18000 mph), the projectile falls towards the Earth in a trajectory that matches the Earth's curvature. So even though the projectile is falling, it never gets any closer to the surface. That means it's in a circular orbit just above the Earth's surface!
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| Circular motion at constant tangential speed
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| An object moving in a circle, even at constant speed, is accelerating because it's always changing direction, and acceleration is the ratio of the change in velocity (speed and direction) with time.
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| For an object moving in a circle at constant speed v, tangent to the circle, the acceleration is directed towards the center of the circle (called centripetal acceleration) and has magnitude aC = v2/R, where R is the radius of the circle.
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| Since a = F/m, then F = ma and the centripetal force required to move an object in a circle of radius R and speed v is FC = maC = mv2/R and the centripetal force is directed towards the center of the circle.
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| One characteristic of circular motion for rotating objects (such as DVD disks) is that all points on the object must go all the way around the circle in the same amount of time. That means that points farther from the point of rotation must travel at greater speeds than points which are closer to the point of rotation. Hence it's frequently easier to think in terms of revolutions per unit time (such as RPM -- revolutions per minute) or angular velocity (radians per second).
Angular velocity is represented by the Greek letter omega ω and its value is the ratio of tangential speed v to the radius R, that is, ω = v/R. If T is the time for one revolution (called the period) of a rotating object, then angular velocity can be determined by ω = 2π/T = 2πf (where f is the frequency of rotation in revolutions per second). That means centripetal acceleration can be expressed in terms of angular velocity: aC = v2/R = Rω2. Similarly, centripetal force can be expressed in terms of angular velocity: FC = maC = mv2/R = mRω2.
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| Find the required tension in the string to swing a 0.20 kg object around in a circle of radius 0.50 m at a speed of 3.0 m/s.
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| FC = maC = mv2/R = (0.20 kg)(3.0 m/s)2/(0.50 m) = 3.6 N
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| A 60-kg woman is riding on a Ferris wheel with radius 10 m which is rotating at 2 revolutions per minute. What is the normal force the seat exerts on the woman at the bottom of the circular motion, and what is the normal force the seat exerts on the woman at the top of the circular motion?
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| The circumference of the Ferris wheel is 2πR = 2(3.14)(10 m) = 62.8 m. The Ferris wheel makes one revolution in 30 s, so the speed of the person is v = 63 m/30 s = 2.1 m/s.
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| The centripetal acceleration of the person is aC = v2/R = (2.1 m/s)2/(10 m) = 0.44 m/s2.
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| The corresponding centripetal force is FC = maC = (60 kg)(0.44 m/s2) = 26 N.
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| A free-body diagram of the woman shows that gravity pulls down on her with a force mg, and the seat pushes up on her with the normal force N.
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| How fast would the Ferris wheel have to turn for her to feel "weightless" at the top of the circle?
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| This is the same as asking (in the last problem) when is the normal force N = mg - mv2/R = 0, so mv2/R = mg, and v = (Rg)1/2 = 17 m/s. That means the time for one revolution would be T = 2πR/v = (63 m)/(17 m/s) = 3.7 s, which is equivalent to saying the Ferris wheel would have to turn about 8 times faster.
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| A car is traveling on a circular road or radius R = 30 m where the coefficient of friction µ = 0.5. What's the fastest speed at which the car can travel without sliding off the road?
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| Side view (looking from behind the car in the direction the car is moving -- the car moving away from us):
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| In the vertical direction, the forces N and mg are balanced, so N = mg.
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| In the horizontal direction, friction is the only force acting and it provides the necessary centripetal force to keep the car moving on a circular road. So the frictional force f = µN = µmg = mv2/R. Solving for v = (µRg)1/2 = ((0.5)(30 m)(9.8 m/s2))1/2 = 12 m/s ≈ 27 miles/hour.
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| A circular road is banked at an angle θ so that a car can make stay on the road at high speed without a lot of friction.
Gravity pulls the car straight down with force mg and the normal force N is tilted at the same angle θ. In the vertical direction, the vertical component of the normal force N cos(θ) balances mg, so N = mg/cos(θ). In the horizontal direction, the horizontal component of the normal force N sin(θ) provides the centripetal force needed to keep the car on a circular road without friction. Hence N sin(θ) = mv2/R. Substituting N = mg/cos(θ), we get mg sin(θ)/cos(θ) = mv2/R. Solving for v = (Rg tan(θ))1/2. If we use R = 30 m, as in the previous example, and choose θ = 10°, we get v = ((30 m)(9.8 m/s2)tan(10°))1/2 = 7.2 m/s.
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| Here is a table showing the speeds rounded to the nearest integer for different angles:
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| Read the first part of Chapter 5, pp. 130-145.
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