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In circular motion, a tangential acceleration can change the magnitude of the velocity but not its direction. Explain your answer.

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Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, centripetal acceleration, or both when: (a) the plate starts to spin faster? (b) The plate rotates at constant angular velocity? (c) The plate slows to a halt?

a. both; b. nonzero centripetal acceleration; c. both

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Problems

At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

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A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is μ S = 0.5 , how far from the axis of rotation can he stand without sliding?

r = 0.78 m

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An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) What is the average angular acceleration in rad/s 2 ? (b) What is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) What is the centripetal acceleration in m/s 2 and multiples of g of this point at full rpm? (d) What is the total distance travelled by a point 9.5 cm from the axis of rotation of the ultracentrifuge?

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A wind turbine is rotating counterclockwise at 0.5 rev/s and slows to a stop in 10 s. Its blades are 20 m in length. (a) What is the angular acceleration of the turbine? (b) What is the centripetal acceleration of the tip of the blades at t = 0 s? (c) What is the magnitude and direction of the total linear acceleration of the tip of the blades at t = 0 s?

a. α = −0.314 rad / s 2 ,
b. a c = 197.4 m / s 2 ; c. a = a c 2 + a t 2 = 197.4 2 + ( −6.28 ) 2 = 197.5 m / s 2
θ = tan −1 −6.28 197.4 = −1.8 ° in the clockwise direction from the centripetal acceleration vector

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What is (a) the angular speed and (b) the linear speed of a point on Earth’s surface at latitude 30 ° N. Take the radius of the Earth to be 6309 km. (c) At what latitude would your linear speed be 10 m/s?

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A child with mass 30 kg sits on the edge of a merry-go-round at a distance of 3.0 m from its axis of rotation. The merry-go-round accelerates from rest up to 0.4 rev/s in 10 s. If the coefficient of static friction between the child and the surface of the merry-go-round is 0.6, does the child fall off before 5 s?

m a = 40.0 kg ( 5.1 m / s 2 ) = 204.0 N
The maximum friction force is μ S N = 0.6 ( 40.0 kg ) ( 9.8 m / s 2 ) = 235.2 N so the child does not fall off yet.

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A bicycle wheel with radius 0.3m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s?

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The angular velocity of a flywheel with radius 1.0 m varies according to ω ( t ) = 2.0 t . Plot a c ( t ) and a t ( t ) from t = 0 to 3.0 s for r = 1.0 m . Analyze these results to explain when a c a t and when a c a t for a point on the flywheel at a radius of 1.0 m.

v t = r ω = 1.0 ( 2.0 t ) m/s a c = v t 2 r = ( 2.0 t ) 2 1.0 m = 4.0 t 2 m / s 2 a t ( t ) = r α ( t ) = r d ω d t = 1.0 m ( 2.0 ) = 2.0 m / s 2 .
Plotting both accelerations gives
Figure shows a linear acceleration in meters per second squared plotted as a function of time in seconds. Centripetal starts at the origin of the coordinate system and grows exponentially with time. Tangential is positive and remains constant with time
The tangential acceleration is constant, while the centripetal acceleration is time dependent, and increases with time to values much greater than the tangential acceleration after t = 1s. For times less than 0.7 s and approaching zero the centripetal acceleration is much less than the tangential acceleration.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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