physics ch6 outline

# physics ch6 outline - Chapter 6 Circular Motion and Other...

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Chapter 6 Circular Motion and Other Applications of Newton’s Laws

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Uniform Circular Motion Uniform circular motion occurs when an object moves in a circular path with a constant speed The velocity vector is always tangent to the path of the object
Changing Velocity in Uniform Circular Motion The change in the velocity vector is due to the change in direction The vector diagram shows f i = + ∆ v v v r r r

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Centripetal Acceleration The acceleration is always perpendicular to the path of the motion and always points toward the center of the circle of motion This acceleration is called the centripetal acceleration. The magnitude of the centripetal acceleration vector is given by 2 C v a r =
Period The period , T , is the time required for one complete revolution The speed of the particle would be the circumference of the circle of motion divided by the period Therefore, the period is defined as 2 r T v π

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Uniform Circular Motion, Force A force, , is associated with the centripetal acceleration The force is also directed toward the center of the circle Applying Newton’s Second Law along the radial direction gives 2 c v F ma m r = = r F r
Example 1 - Conical Pendulum A 50 g ball is suspended from a string of length L = 80 cm. The ball revolves with constant speed in a horizontal circle of radius r. If θ = 12.2 0 ,what is the speed of the ball. The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction ∑F y = 0 → T cos θ = mg ∑F x = T sin θ = m a c

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Example 2 - Motion in a Horizontal Circle A 0.25 kg mass is attached to the end of a cord 65 cm long. The ball is whirled in a horizontal circle. If the cord can withstand a maximum tension of 50 N, what is the maximum speed at which the ball can be whirled before the cord breaks? The centripetal force is supplied by the tension
Example 3 - Horizontal (Flat) Curve A 1000-kg car moving on a flat, horizontal road negotiate a curve as shown. If the radius of the curve is 50 m and the coefficient of static friction between the tires and the road is 0.53, find the maximum speed the car can have and still make the turn successfully The force of static friction supplies the centripetal force: μ s mg =ma c a c = v 2 /R The maximum speed at which the car can negotiate the curve is Note, this does not depend on the mass of the car s v gr μ =

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