Topic 2: Mechanics 2.4 Uniform circular motion

2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle.

2.4.2 Apply the expressions for centripetal acceleration.

2.4.3 Identify the force producing circular motion in various situations. Examples include friction of tires on turn, gravity on planet/moon, cord tension.

2.4.4 Solve problems involving circular motion.

Topic 2: Mechanics2.4 Uniform circular motion

What force must be applied to Helen to keep her moving in a circle?

How does it depend on the Helen’s radius r?

How does it depend on Helen’s velocity v?

How does it depend on Helen’s mass m?


On the next pass, however, Helen failed to clear the mountains.

r

vm

Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle.

A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v.

Observe that the velocity vector is always tangent to the circle.

Note that the magnitude of the velocity vector is NOT changing.

Note that the direction of the velocity vector IS changing.

Thus, there is an acceleration, even though the speed is not changing!


x

y

r

v

r bluev red

Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle.

In order to find the direction of the acceleration (a = ∆v/∆t ) we observe two nearby snapshots of the particle:

The direction of the acceleration is gotten from ∆v = v2

– v1 = v2 + (-v1):

The direction of the acceleration is toward the center of the circle- you must be able to sketch this.


x

yr bluev red

v1

v2

v1

v2

∆vv1

v2

-v1

∆v-v1

Apply the expressions for centripetal acceleration.

Centripetal means center-seeking.How does centripetal acceleration ac depend on r and v?

We define the centripetal force Fc:

Picture yourself as the passenger in a car that is rounding a left turn:

The sharper the turn, the harder you and your door push against each other. (Small r = big Fc.)

The faster the turn, the harder you and your door push against each other. (Big v = big Fc.)


Fc = mac centripetal force

Fc

PRACTICE: For each experiment A and B, label the control, independent, and dependent variables.



Fc

ac

v

r

A B

Fc

ac

v

r

Fc

ac

v

r

Fc

acv

r

CONTROL: rINDEPENDENT: v

DEPENDENT: Fc and ac

CONTROL: vINDEPENDENT: r

DEPENDENT: Fc and ac

manipulated

No change

responding

no change

manipulated

responding


We know the following things about ac:

If v increases, ac increases.

If r increases, ac decreases.

From dimensional analysis we have

What can we do to v or r to “fix” the units?

This is the correct one!


ac =vr

ac =vr

ms2

=? 1

sm/sm

=?

ac = v2/r centripetal acceleration

ac =v2

r m

s2=? m

s2=?

first guess formula

m2/s2

m



ac = v2/r centripetal acceleration

EXAMPLE: A 730-kg Smart Car negotiates a 30. m radius turn at 25. m s-1. What is its centripetal acceleration and force? What force is causing this acceleration?

SOLUTION:ac = v2/r = 252/30 = 21 m s-2.

Fc = mac = (730)(21) = 15000 n.

The centripetal force is caused by the friction force between the tires and the pavement.

Fc = mac centripetal force


The period T is the time for one complete revolution.

One revolution is one circumference C = 2r. Therefore v = distance / time = 2r/T.Thus v2 = 42r2/T2 so that

ac = v2/r

= 42r2/T2r

= 42r/T2.


ac = v2/r centripetal accelerationac = 42r/T2



ac = v2/r centripetal accelerationac = 42r/T2

EXAMPLE: Albert the 2.50-kg physics cat is being swung around your head by a string harness having a radius of 3.00 meters. He takes 5.00 seconds to complete one fun revolution. What are ac and Fc? What is the tension in the string?

SOLUTION:ac = 42r/T2 = 42(3)/(5)2 = 4.74 m s-2.

Fc = mac = (2.5)(4.74) = 11.9 n.

The tension is causing the centripetal force, so the tension is Fc = 11.9 n.

Albert the Physics Cat

EXAMPLE: Dobson is watching a 16-pound bowling ball being swung around at 50 m/s by Arnold. If the string is cut at the instant the ball is next to the ice cream, what will the ball do?

(a)It will follow path A and strike Dobson's ice cream.

(b)It will fly outward along curve path B.

(c)It will fly tangent to the original circular path along C.

Solve problems involving circular motion.


B

A

C

EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone.

(a) Sketch in the forces acting on the mass.

SOLUTION:

The ONLY two forces acting on the mass are its weight W and the tension in the string T.

Don’t make the mistake of drawing Fc into the diagram.

Fc is the resultant of T and W, as the next questions will illustrate.



1.50 m

W

T


(b) Draw a FBD in the space provided. Then break down the tension force in terms of the unknown tension T.

SOLUTION:

Tx = T cos

= T cos 60° = 0.5T.

Ty = T sin

= T sin 60° = 0.87T.



W

1.50 m

W

TT

Tx

Ty

EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60.°, with the mass traveling the base of the cone.

(c) Find the value of the components of the tension Tx and Ty.

SOLUTION:

Note that Ty = mg = 3.0(9.8) = 29.4 n.

But Ty = 0.87T.

Thus 29.4n = 0.87T so that T = 33.8 n.Tx = 0.5T = 0.5(46) = 23 n.



W

T

Tx

Ty


(d) Find the speed of the mass as it travels in its circular orbit.

SOLUTION:

The center of the UCM is here:Since r / 1.5 = cos 60°,r = 0.75 m.

Fc = Tx = 23 n.

Thus Fc = mv2/r

23 = 3v2/0.75

v = 2.4 m s-1.



1.50 m

r

EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of RE = 6400000 m, find the speed of the ball.

SOLUTION:

The ball is traveling in a circle of radius r = 6408850 m.

Fc is caused by the weight of the ball so that

Fc = mg = (0.5)(10) = 5 n.

Since Fc = mv2/r we have

5 = (0.5)v2/6408850

v = 8000 m s-1!



http://de.wikipedia.org/w/index.php?title=Datei:Everest_North_Face_toward_Base_Camp_Tibet_Luca_Galuzzi_2006.jpg&filetimestamp=20070320004704

EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). How long will it take the ball to return to Everest?

SOLUTION:

We want to find the period T.We know that v = 8000 m s-1.

We also know that r = 6408850 m.Since v = 2r/T we have T = 2r/v T = 2(6408850)/8000 T = (5030 s)(1 h / 3600 s)

= 1.40 h.



http://de.wikipedia.org/w/index.php?title=Datei:Everest_North_Face_toward_Base_Camp_Tibet_Luca_Galuzzi_2006.jpg&filetimestamp=20070320004704

EXAMPLE: Explain how an object can remain in orbit yet always be falling.

SOLUTION:

Throw the ball at progressively larger speeds.

In all instances the force of gravity will draw the ball toward the center of the earth.

When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface.

The ball is effectively falling around the earth!



Topic 2: Mechanics 2.4 Uniform circular motion

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