Circular Motion 2nd day

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Circular Motion 2nd day
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-The diagram below depicts the car on the left
side of the circle. -the net force acting upon
the object is directed inwards                    
                    
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Free-body diagram
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• Sample Problem 1
• 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. Sample Problem 2
• The coefficient of friction acting upon a 900-kg
  car is 0.850. The car is making a 180-degree turn
  around a curve with a radius of 35.0 m. Determine
  the maximum speed with which the car can make the
  turn.

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Fgrav m g Fgrav Fnorm 9000 N
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• Since the force of friction is the only
  horizontal force, it must be equal to the net
  force acting upon the object.

Thus, the force of friction is 3600 N.
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• Substituting 3600 N for Ffrict and 9000 N for
  Fnorm yields a coefficient of friction of 0.400.
•  

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• Known Information
• m 900 kg "mu" 0.85 (coefficient of friction)
• R 35.0 m
• Requested Information
• v ??? (the minimum speed would be the speed
  achieved with the given friction coefficient)

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Fgrav m g

• Fgrav Fnorm 9000 N

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Fnet ma

Substituting the known values for a and R into
this equation and solving algebraically yields a
maximum speed of 16.4 m/s.
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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.                           
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2. A 1.5-kg bucket of water is whirled in a
circle with a radius of 1.0 m. At the bottom of
the 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.                    
      
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• The end centripetal motion