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Circular Motion

Circular Motion. Uniform Circular Motion. An object that moves in a circle at a constant speed, v . The magnitude of the velocity remains the same but the direction is changing so the velocity changes. Acceleration .

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Circular Motion

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  1. Circular Motion

  2. Uniform Circular Motion • An object that moves in a circle at a constant speed, v. • The magnitude of the velocity remains the same but the direction is changing so the velocity changes

  3. Acceleration • Acceleration is the change in velocity that occurs over time. Since changing direction changes the velocity the object moving in a circle is constantly accelerating.

  4. Centripetal acceleration • Centripetal Acceleration (center seeking) • Also known as radial acceleration aR (directed along the radius of the circle) • aR = v2 /r • V is the linear speed, • R is the radius of the circle. • The direction of the acceleration is toward the center of the circle.

  5. Frequency and Period • Time Period, T • The time period of a circular motion is the time taken for one revolution. • Rotational Frequency, f • The rotational frequency of a circular motion is the number of revolutions per unit time. • Thus the relation between time period and frequency is T = 1/f

  6. d = One revolution is 2 π r • v = d/t and t = T So v = 2 π r T

  7. Newton’s 2nd law (dynamics) • Σ FR = m aR = (mv2 )/r • Since the acceleration is directed toward the center, and force and acceleration are in the same direction, the force is also directed toward the center.

  8. Centrifugal force? • Not a real force. • Because 3rd law says forces must occur in pairs this is a made up force that explains the outward force that is felt when you go in a circle. • In order to move in a circle a force must be applied to keep it in a circle. Otherwise the object will keep going in the direction that it wants to go without that force. THERE IS NO OUTWARD pushing FORCE (no centrifugal force)

  9. Sum of the forces • If working in the vertical direction or at an angle, other forces must be considered as well.

  10. Maximum Speed on Banked Roadway

  11. Game Plan • Draw a free body diagram • Write down the givens • Write down what you don’t know and what you need. • Resolve any forces that are at angles into the components so that you are only dealing with x and y forces. • Determine the magnitude of any known forces (e.g. the force of gravity and label on the diagram • Use the circular motion equations to find the unknowns.

  12. examples Sample Problem #1 • A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car. • Known Information: • m = 900 kg v = 10.0 m/s R = 25.0 m • Requested Information:a = ???? Fnet = ???? • a = (v2)/R • a = ((10.0 m/s)2)/(25.0 m) = (100 m2/s2)/(25.0 m) a = 4 m/s2 • Fnet = m*a Fnet = (900 kg)*(4 m/s2) Fnet = 3600 N

  13. more • Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters. • A 900-kg car makes a 180-degree turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car.

  14. 1. A 1.5-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.0 m. At the top of the circular loop, the speed of the bucket is 4.0 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop. m = 1.5 kg a = ________ m/s/s Fnet = _________ N 2. A 1.5-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.0 m. At the bottom of the circular loop, the speed of the bucket is 6.0 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop. m = 1.5 kg a = ________ m/s/s Fnet = _________ N

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