Chapter 5: Uniform Circular Motion - PowerPoint Presentation

Slide1

Chapter 5: Uniform Circular MotionSlide2

5-1 Uniform Circular Motion

Uniform Circular Motion

: The motion of an object traveling at a constant (uniform) speed on a circular pathSlide3

5-1 Uniform Circular Motion

Since we are dealing with object moving in a circle, it is convenient to talk about the

period

of the motion

Period (T)

= the time required to travel once around the circle (one complete revolution.Slide4

5-1 Uniform Circular Motion

Period (T)

= the time required to travel once around the circle (one complete revolution.

The distance around the circle is just the circumference of the circle (2

π

r)

Speed (v) =

Notice v is scalar – it gives the speed only (no direction)

 Slide5

5-1 Uniform Circular Motion

The blades on the

turbines at the Wild

Horse Wind Farm in

eastern WA turn at a rate of 16 revolutions/minute. Each blade is approx. 39m long. Compare the speed of the tip of the blade to the speed of the part of the blade that is 1m from the center.Slide6

5-1 Uniform Circular Motion

First, find the time it

Takes for one revolution.

f= 16 rev/minute

= 16 rev/60 sec

= 0.267 rev/secT = 1/f = 3.75 secondsSlide7

5-1 Uniform Circular Motion

Next, find the distance

For one revolution

r=39m

Distance = 2

π

r = 245m

v = distance/time = 245m/3.75s = 65.35 m/s speed of the tipSlide8

5-1 Uniform Circular Motion

Next, find the distance

For one revolution

r=1m

Distance = 2

π

r = 6.3m

v = distance/time = 6.3m/3.75s = 1.68 m/s speed of the inner part of the blade.Slide9

5-1 Uniform Circular Motion

For uniform circular motion, the speed is constant, but the velocity is not because the direction is changing continually!

At any instant, the velocity

v

ector is always tangent to

the circle.Slide10

Check Your Understanding

1. A tube is been placed upon the table and shaped into a three-quarters circle. A golf ball is pushed into the tube at one end at high speed. The ball rolls through the tube and exits at the opposite end. Describe the

path

of the golf ball as it exits

the

tube.Slide11

Check Your Understanding

While the ball is in the tube, the tube exerts a force on the ball which causes the ball to accelerate in a circle. Once the ball leaves the tube, there are no forces acting on it. The ball will continue to move in a straight line.Slide12

Check Your Understanding

Draw the velocity vector at each point (A,B,C) in the object’s motion.Slide13

Centripetal Acceleration

Draw the velocity vector at each point (A,B,C) in the object’s motion.

If the object is accelerating (velocity is changing), then by Newton’s 2

nd

Law, it must have a net force acting on it.

vSlide14

Recall from the bowling ball lab that a net inward

force

was required to move the ball in a circle.

The force and acceleration point in the same direction. Therefore, the object’s

acceleration

is towards that center of the circle at any given instant.

v

v

v

a

a

aSlide15

Centripetal Acceleration

We call this acceleration the centripetal acceleration.

a

c

=

The centripetal acceleration vector always points toward the center of the circle, and changes direction as the object moves.Slide16

Example 1

The bobsled track at Lake Placid, NY (site of the 1980 winter

O

lympics), contains turns with radii of 33m and 24m. Find the centripetal acceleration at each turn for a speed of 34m/s. How many “g’s” is this?

a

c

=

r=33m ac

=35m/s

2

=3.6g

r=24m a

c

=48m/s

2

=4.9g

 Slide17

Geometry Review

How do you find the length of an arc, s,?

θ

s

r

s

=r

θ

But

θ

must be in

radians!Slide18

Assignment

P. 156 #1, 3, 5, 7, 8, 9, 10