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These are isosceles triangles; therefore, ∆v/ v = v∆t / r

Uniform Circular Motion is the motion of an object traveling at a constant (uniform) speed on a circular path. v = 2πr/T The velocity of uniform circular motion is the circumference divided by the period of one revolution.

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These are isosceles triangles; therefore, ∆v/ v = v∆t / r

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  1. Uniform Circular Motion is the motion of an object traveling at a constant (uniform) speed on a circular path.

  2. v = 2πr/TThe velocity of uniform circular motion is the circumference divided by the period of one revolution.

  3. Ex. 1 - The wheel of a car has a radius of 0.29 m and is being rotated at 830 revolutions per minute on a tire-balancing machine. Find the speed in m/s at which the outer edge of the wheel is moving.

  4. In uniform circular motion the magnitude of the velocity vector is constant. The direction; however, is constantly changing. This change is an acceleration which is called “centripetal acceleration”.

  5. These are isosceles triangles; therefore, ∆v/ v = v∆t / r

  6. ∆v/ v = v∆t / rThis can be solved for ∆v/∆t, to show that ac = v2/r.

  7. The direction of the vector quantity ac is toward the center of the circle.

  8. Ex. 3 - The bobsled track at the Olympics contained turns with radii of 33 m and 24 m. Find the centripetal acceleration for each turn for a speed of 34 m/s. Express the answers as multiples of g = 9.8 m/s2.

  9. An object in uniform circular motion is constantly being accelerated. This is a non-inertial frame of reference; and an object in uniform circular motion can never be in equilibrium.

  10. The force that pulls an object into a circular path is called a centripetal force. F = ma, so Fc = mac; therefore, Fc = mv2/r

  11. Ex. 5 - A model airplane has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. Find the tension T in the guideline (length = 17 m) for speeds of 19 and 38 m/s.

  12. When a car moves around an unbanked curve, the static friction of the tires on the road provides the centripetal force that keeps the car from sliding, at least, we hope so......

  13. Ex. 7 - Compare the maximum speeds at which a car can safely negotiate an unbanked turn (radius = 50.0 m) in dry weather (coefficient of static friction = 0.900) and icy weather (coefficient of static friction = 0.100).

  14. Banked curves on roadways provide a centripetal force to help the car into a curved path. The inward pointing component of the normal force provides the centripetal force.

  15. FC = FNsin q = mv2 / rAlso, FNcos q = mg(vertical component must equal weight)

  16. FNsin q = mv2/r ---------------------FNcos q = mg

  17. tan q = v2 /rg

  18. Ex. 8 - The turns at the Daytona 500 have a maximum radius (at the top) of r = 316 m and are banked steeply, with q = 31°. Suppose these turns were frictionless. At what speed would the race cars have to travel around them?

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