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This guide explores the concepts of uniform circular motion, including definitions of uniform motion, velocity direction, and acceleration in circular paths. It covers the magnitude of acceleration, effects of changing speed and radius, and the significance of centripetal force. Using practical examples, such as swinging a bucket of water, we illustrate the normal forces acting on the water inside the bucket at different points in its circular path. The guide serves as a comprehensive resource for students to grasp the foundational principles of circular motion in physics.
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Uniform circular motion • What’s uniform? • What direction is the velocity? • What direction is the acceleration? • What’s the magnitude of the acceleration?
What is the acceleration? In one second: v
In one second: ac What is the acceleration?
What if the speed is doubled? In one second:
v 2v In one second: ac
What if the radius is halved? In one second:
What if the radius is halved? In one second:
What if the radius is halved? In one second:
What if the radius is halved? In one second: ac
What if the radius is halved? In one second: r r/2
Centripetal Acceleration • acceleration for objects moving in a circle at constant speed • perpendicular to the velocity, which points towards the center of the circle • ac = v2/r
TS,B WE,B Free-body diagram: If you can’t label the force, it probably doesn’t belong on your diagram! Ball ??
TS,B WE,B Free-body diagram: Ball
Free-body diagram: TS,B Ball WE,B
Free-body diagram: Ty Ball Tx WE,B
What direction is the net force on the ball? Ty Ball Tx WE,B
What direction is the net force on the ball? Ty Ball Tx WE,B a
For an object moving in a circle: • the acceleration points towards the center of the circle, and • the NET FORCE points towards the center of the circle. • We give a special name to this NET FORCE; we call it the CENTRIPETAL FORCE.
Which force is the NET FORCE? Ty Ball Tx WE,B a
What about the centrifugal force? • There is no such force, regardless of what Mr. Wizard says.
Bucket of water problem • You swing a bucket of water (m=1kg) in a vertical circle at constant speed, which we’ll estimate. What is the normal force by the bottom of the bucket on the water at:a. the top of the circle, andb. the bottom of the circle.
WE,W FBD at top water
NB,W What else? water WE,W
v a a NB,W What direction is a? water WE,W Why doesn’t the water fall on Greg’s head?
v a NB,W But, how big is a? a water WE,W Why doesn’t the water fall on Greg’s head?
1 m 2 p r Assume Dt = 1 sec; v = ≈ 6 m/s D t Centripetal acceleration • aC = v2/r • Assume r = 1 meter ≈ 36 m/s2 • need to find v
NB,W Question a = aC water WE,W
What is the acceleration? In one second: v
In one second: ac What is the acceleration?
NB,W Apply Newton’s 2nd Law a = 36 m/s2 water WE,W = m g = 10 N What’s wrong with this freebody diagram? Fnet = (1 kg) (36 m/s2) = 36 N
NB,W What if we swing a little slower? a water WE,W
NB,W What if we swing a little slower? a water WE,W
NB,W What has to change on the FBD? a water WE,W
What has to change on the FBD? a water NB,W WE,W
NB,W What changes at the bottom? a water WE,W
a NB,W WE,W At the bottom: water
WE,W At the bottom: a NB,W water
WE,W Apply Newton’s 2nd Law a = 36 m/s2 NB,W water = m g = 10 N How big is the normal force? Fnet = (1 kg) (36 m/s2) = 36 N