Uniform Circular Motion Power Point

Uniform Circular Motion

©2009 by Goodman & Zavorotniy

Kinematics of UCM

Topics of Uniform Circular Motion (UCM)Click on the topic to go to that section

Dynamics of UCM

Vertical UCMBuckets of WaterRollercoastersCars going over hills and through valleys

Horizontal UCMUnbanked CurvesBanked CurvesConical Pendulum

Period, Frequency, and Rotational Velocity

Key Terms and Equations

Centripetal Force Equations:

aR = v2/r

FC = mv2/r

Period, Frequency, & Rotational Velocity Equations:

T = t/n = 1/ff = n/t = 1/Tv = 2πr/T = 2πrf

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Kinematics of UCM

Kinematics of Uniform Circular Motion

Uniform circular motion: motion in a circle of constant radius at constant speed

Instantaneous velocity is always tangent to circle.

This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.


Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that we get two similar triangles.


A B

C

rr

A

B

rr

C

Δθ

Δl

Δθ

If displacement is equal to velocity multiplied by time, then vt is the displacement covered in time t.


θr

r

vt v1

v2θ

v1

v2 Δv

During that same time, velocity changed by an amount, Δv.

These are similar triangles because the angles are all congruent, so the sides must be in proportion.

Δv vt v r

Δv v2 t r

v2 r

=

=

a = v1

v2Δv

θ

vt

rr

θ


These are similar triangles because the angles are all congruent, so the sides must be in proportion.

v1 v2

Δv

θ

vt

rr

θ


That's the magnitude of the acceleration.

Δv vt v r

Δv v2 t r

v2 r

=

=

a =

v1

v2

θ


The change in velocity, Δv, shows the direction of the acceleration. In the picture, you can see Δv points towards the center of the circle.

v1

v2 Δv

θ

Transpose v2 to see the vector addition.

This acceleration is called the centripetal, or radial, acceleration.

It direction is towards the center of the circle.

It's magnitude is given by


a = v2/r

1 Is it possible for an object moving with a constant speed to accelerate? Explain.

A No, if the speed is constant then the acceleration is equal to zero.

B No, an object can accelerate only if there is a net force acting on it.

C Yes, although the speed is constant, the direction of the velocity can be changing.

D Yes, if an object is moving it is experiencing acceleration.

2 Consider a particle moving with constant speed such that its acceleration of constant magnitude is always perpendicular to its velocity.

A It is moving in a straight line.B It is moving in a circle.C It is moving in a parabola.

D None of the above is definitely true all of the time.

3 An object moves in a circular path at a constant speed. Compare the direction of the object's velocity and acceleration vectors.

A Both vectors point in the same direction.B The vectors point in opposite directions.C The vectors are perpendicular.

D The question is meaningless, since the acceleration is zero.

4 What type of acceleration does an object moving with constant speed in a circular path experience?

A free fallB constant accelerationC linear accelerationD centripetal acceleration

5 An object is traveling with a velocity of 6.0 m/s in a circular path whose radius is 4.0m. What is the magnitude of its centripetal acceleration?

A 4 m/s2

B 9 m/s2

C 6 m/s2

D 1.5 m/s2

E .67 m/s2

6 An object is traveling with a velocity of 6.0 m/s in a circular path. Its acceleration is 3.0 m/s2. What is the radius of its path?

A 6 m

B 1 m

C 2 m

D 16 m

E 12 m

7 An object is traveling with a velocity in a circular path whose radius is 65 m. Its acceleration is 3.0 m/s2. What is its velocity?

A 11.87 m/s

B 15.36 m/s

C 19.74 m/s

D 13.96 m/sE 195 m/s


Period, Frequency, andRotational Velocity

Period

The time it takes for an object to complete one trip around a circular path is called its Period.

The symbol for Period is "T"

Periods are measured in units of time; we will usually use seconds (s).

Often we are given the time (t) it takes for an object to make a number of trips (n) around a circular path. In that case,

T = t/n

8 If it takes 50 seconds for an object to travel around a circle 5 times, what is the period of its motion?

A 5 sB 10 sC 15 sD 20 sE 25 s

9 If an object is traveling in circular motion and its period is 7.0s, how long will it take it to make 8 complete revolutions?

A 56 sB 54 sC 63 sD 58 sE 48 s

FrequencyThe number of revolutions that an object completes in a given amount of time is called the frequency of its motion.

The symbol for frequency is "f"

Frequency are measured in units of revolutions per unit time; we will usually use 1/seconds (s1). Another name for s1 is Hertz (Hz). Frequency can also be measured in revolutions per minute (rpm), etc.

Often we are given the time (t) it takes for an object to make a number of revolutions (n). In that case,

f = n/t

10 An object travels around a circle 50 times in ten seconds, what is the frequency (in Hz) of its motion?

A 25 Hz

B 20 Hz

C 15 Hz

D 10 Hz

E 5 Hz

11 If an object is traveling in circular motion with a frequency of 7.0 Hz, how many revolutions will it make in 20 s?

A 120

B 145

C 130

D 140

E 150

Period and Frequency

Since T = t/n

and f = n/t

then T = 1/f

and f = 1/T

12 An object has a period of 4.0s, what is the frequency of its motion (in Hertz)?

A 1/16 Hz

B 1/8 Hz

C 1/4 Hz

D 1/2 Hz

E 2 Hz

13 An object is revolving with a frequency of 8.0 Hz, what is its period (in seconds)?

A 1/8 s

B 1/4 s

C 1/2 sD 2 sE 4 s

Rotational VelocityEach trip around a circle, an object travels a length equal to the circle's circumference.

The circumference of a circle is given by: C = 2πr

The time it takes to go around once is the period, T.

And the object's speed is given by s = d/t

So the speed must be:

s = C/T = 2πr/T

Rotational VelocityA velocity must have a magnitude and a direction.

The magnitude of an object's instantaneous velocity is its speed. So for an object in uniform circular motion, the magnitude of its velocity is:

v = C/T = 2πr/T

If an object is in uniform circular motion, the direction of its velocity is tangent to its circular motion.

So v = 2πr/T tangent to the circle

14 An object is in circular motion. The radius of its motion is 2.0m and its period is 5.0s. What is its velocity?

A 1.98 m/s

B 3. 57 m/s

C 4.36 m/s

D 3.25 m/sE 2.51 m/s

15 An object is in circular motion. The radius of its motion is 2.0 m and its velocity is 20 m/s. What is its period?

A 0.38 s

B 0.63 s

C 0.78 s

D 0.89 sE 1.43 s

16 An object is in circular motion. The period of its motion is 2.0 s and its velocity is 20 m/s. What is the radius of its motion?

A 7.87 m

B 3.56 m

C 5.61 mD 6.36 m

E 5.67 m

Rotational VelocitySince f = 1/T, we can also determine the velocity of an object in uniform circular motion by the radius and frequency of its motion.

v = 2πr/T and f = 1/T so

v = 2πrf

Of course the direction of its velocity is still tangent to its circular motion.

So v = 2πrf tangent to the circle

17 An object is in circular motion. The radius of its motion is 2.0 m and its frequency is 8.0 Hz. What is its velocity?

A 100.53 m/s

B 106.89 m/s

C 97.93 m/s

D 102.23 m/sE 103.39 m/s

18 An object is in circular motion. The radius of its motion is 2.0 m and its velocity is 30 m/s. What is its frequency?

A 4.12 Hz

B 2.82 Hz

C 2.39 Hz

D 3.67 Hz

E 1.78 Hz

19 An object is in circular motion. The frequency of its motion is 7.0 Hz and its velocity is 20 m/s. What is the radius of its motion?

A 0.45 m

B 2.08 mC 0.33 m

D 1.22 m

E 1.59 m


Dynamics of UCM

Dynamics of Uniform Circular MotionFor an object to be in uniform circular motion, there must be a net force acting on it.

We already know the acceleration, so we can write the force:

We can see that the force must be inward by thinking about a ball on a string:

Dynamics of Uniform Circular Motion

Force on ballexerted bystring

Force on handexerted bystring

There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.

If the centripetal force vanishes, the object flies off tangent to the circle.

Dynamics of Uniform Circular Motion

This happens.

This does NOT happen.

This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular.

Curved Paths

Centrifugation

A centrifuge works by spinning very fast. This means there must be a very large centripetal force. The object at A would go in a straight line but for this force; as it is, it winds up at B.

20 What force is needed to make an object move in a circle?

A kinetic frictionB static frictionC centripetal forceD weight

21 When an object experiences uniform circular motion, the direction of the net force is

A in the same direction as the motion of the object

B in the opposite direction of the motion of the object

C is directed toward the center of the circular path

D is directed away from the center of the circular path

22 A car with a mass of 1800 kg goes around an 18 m radius turn at a speed of 35 m/s. What is the centripetal force on the car?

23 A 75 kg mass is attached to the end of a 5.0 m long metal rod string which rotates in a horizontal circular path. If the maximum force that the rod can withstand is 8500 N. What is the maximum speed that the mass can attain without breaking the rod?


Vertical UCM

Car on a hilly road....

A car is traveling at a velocity of 20 m/s.

The driver of the car has a mass of 60 kg.

The car is located at the bottom of a dip in the road. The radius of the dip is 80m.

What is the apparent weight of the driver (the normal force supplied by the seat of the car to support him) at the bottom of the dip?

This is a sketch of the problem.

What do I do next?

1. Draw a free body diagram

2. Indicate the direction of acceleration

v = 20 m/sm = 60 kgr = 80 m

mg

FNa

Next,

3. Draw axes with one axis parallel to acceleration

v = 20 m/sm = 60 kgr = 80 m

mg

FNa

y

x

The dotted lines represent axes with one axis parallel to acceleration

All forces are parallel or perpendicular to the axes, so I don't have to resolve any vector into components

v = 20 m/sm = 60 kgr = 80 m

mg

FNa

y

x

x direction

∑F = ma0 = 0

y direction

∑F = maFN mg = ma FN = mg + ma FN = m (g + a)

4. Apply Newton's Second Law along each axis.

While I don't know "a", I do know "v" and "r". What's my next step?

5. Substitute a = v2/r

FN = m (g + v2/r)

v = 20 m/sm = 60 kgr = 80 m

mg

FNa

x

v = 20 m/sm = 60 kgr = 80 m

6. The last step.

Substitute numbers.

FN = m (g + v2/r)

FN = (60 kg) ((9.8 m/s2 + (20 m/s)2/(80m))

FN = (60 kg) (9.8 m/s2 + 5 m/s2)

FN = (60 kg) (14.8 m/s2 )

FN = (60 kg) (14.8 m/s2 )

FN = 890 N

How does this compare to his weight on the flat part of the road?

On the flat road the driver's weight (the normal force from the seat) is just mg.

FN = mg

FN = (60kg)(9.8 m/s2) = 590 N

Apparent Weight

Flat road Bottom of dip (r = 80m)

590 N 890 N

The gravitational force on the driver (mg) doesn't change, but his apparent weight (FN) does.

Is there a situation where he will appear weightless?

FN

mg

v = 20 m/sm = 60 kgr = ∞

At what velocity must a car drive over a hill if the driver (and the car for that matter) are to appear weightless?


The radius of the hill is 80m.

m = 60 kgr = 80 m

This is a sketch of the problem.

What do I do next?

1. Draw a free body diagram

2. Indicate the direction of acceleration

mg

FN

a

m = 60 kgr = 80 m This is the free body diagram

(let's start out with FN, knowingwe'll eventually let it equal zero)

Next,

3. Draw axes with one axis parallel to acceleration

mg

FN

a

y

x

The dotted lines represent axes with one axis parallel to acceleration

m = 60 kgr = 80 m

x direction

ΣF = ma0 = 0

y direction

ΣF = maFN mg = ma FN = mg ma FN = m (g a)

4. Apply Newton's Second Law along each axis.

But I'd like to find the velocity that the car must be going for the driver (and car) to appear weightless. How?

5. Substitute a = v2/r

FN = m (g v2/r)

mg

FN

a

y

x

m = 60 kgr = 80 m

6. The last step, when does FN = 0

When the product of two variables is zero, then one of them must be zero. Since m is not zero, the only way Fn can equal zero is...

FN = m (g v2/r)

0 = (60 kg) (g v2/r)

0 = (g v2/r)

v2/r = g

v = (gr)1/2

v = ((9.8m/s2)(80m))1/2

v = 28 m/s

mg

FN

a

y

x

m = 60 kgr = 80 m

Practice Problem:A car is traveling at a velocity of 15 m/s.


The car is located at the top of a hill in the road. The radius of the hill is 45 m.

What is the apparent weight of the driver (the normal force supplied by the seat of the car to support him) at the top of the hill?

240 N

24 A car is going over the top of a hill whose curvature approximates a circle of radius 175 m. At what velocity will the occupants of the car appear to weigh 10% less than their normal weight (FN = 0.9mg)?

A 13.1 m/s

B 14. 7 m/s

C 13.9 m/s

D 14.2 m/sE 12.7 m/s

25 A car is going through a dip in the road whose curvature approximates a circle of radius 175 m. At what velocity will the occupants of the car appear to weight 10% more than their normal weight?

A 9.8 m/s

B 12.7 m/s

C 11.9 m/s

D 13.1 m/sE 14.5 m/s

26 The occupants of a car traveling at a speed of 25 m/s note that on a particular part of a road their apparent weight is 20% higher than their weight when driving on a flat road.

Is that part of the road a hill, or a dip?

A HillB Dip

27 The occupants of a car traveling at a speed of 25 m/s note that on a particular part of a road their apparent weight is 20% higher than their weight when driving on a flat road.

What is the vertical curvature of the road?

A 345.67 mB 298.74 mC 276.91 mD 399.35 m

E 318.88 m

Buckets and Rollercoasters...

A bucket of water is being spun is a vertical circle of radius 0.80m. What is the smallest velocity which will result in the water not leaving the bucket?

What do you know and what are you looking for?

r = 0.80 m

g = 9.8 m/s2 downward

v = ?

r = 0.80 m g = 9.8 m/s2 downward vminimum = ?

mg

T

a


mg

T

ΣF = ma


mg

T

ΣF = mv2/r


mg

T

ΣF = mv2/r


ΣF = ma

T + mg = m(v2/r)

T = m(v2/r) mg

T = m(v2/r g)

T = 0 for minimum v

v2/r g = 0

v2/r = g

v = (gr)1/2 = ((9.8)(0.8))1/2 = 2.8 m/s

T

Assuming a constant speed, and that the mass of the bucket is 2.5kg, what will the tension in the string be at the bottom of the circle?

r = 0.80 m g = 9.8 m/s2 downward

mg

ΣF = mv2/r

T

Assuming a constant speed, and that the mass of the bucket is 2.5kg, what will the tension in the string be at the bottom of the circle?

ΣF = ma

T mg = ma

T mg = m(v2/r)

T = m(v2/r) + mg

T = m(v2/r + g)

T = 2.5kg (9.8 m/s2 + 9.8 m/s2)

T = 2.5kg (19.6 m/s2)

T = 490 N

r = 0.80 m g = 9.8 m/s2 downward

Practice Problem:A bucket of water is being spun is a vertical circle of radius 1.2 m. What is the smallest velocity which will result in the water not leaving the bucket?

3.43 m/s

Assuming a constant speed, and that the mass of the bucket is 1.3 kg, what will the tension in the string be at the bottom of the circle?

25.48 N

28 A ball is attached to the end of a string. It is swung in a vertical circle of radius 10 m. What is the minimum velocity that the ball must have to make it around the circle?

A 2.8 m/sB 9.9 m/s

C 7.5 m/s

D 2.1 m/sE 3.9 m/s

29 A ball is attached to the end of a string. It is swung in a vertical circle of radius 2.25 m. What is the minimum velocity that the ball must have to make it around the circle?

A 3.87 m/s

B 4.34 m/s

C 4.69 m/s

D 5.12 m/sE 5.39 m/s

30 A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with track at the top of the loop, what is the minimum value for its centripetal acceleration at this point?

A g downwardB 0.5g downwardC g upwardD 2g upward

31 A roller coaster car (mass = M) is on a track that forms a circular loop (radius = r) in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its speed at that point?

A rgB (rg)1/2

C (2rg)1/2

D (0.5rg)1/2

32 A pilot executes a vertical dive then follows a semicircular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is

A less than mg and pointing upB less than mg and pointing downC more than mg and pointing upD more than mg and pointing down


Horizontal UCM

When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

Banked and Unbanked Curves


If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

As long as the tires do not slip, the friction is static.

If the tires do start to slip, the friction is kinetic, which is bad in two ways:

• The kinetic frictional force is smaller than the static.

• The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.


Unbanked Curves

A car is going around a track with a velocity of 20 m/s. The radius of the track is 150 m. What is the minimum coefficient of static friction that would make that possible?

What do you know and what are you looking for?

v = 20 m/s

r = 150 m

μ = ?

Front View(the car heading towards you)

r

Top View

v = 20 m/s r = 150 m μ = ?

Unbanked Curves

FN

fs

mg

Front View(the car heading towards you)

v

ar

Top View

v = 20 m/s r = 150 m μ = ?

Unbanked Curves

FN

fs

mg

vertical

radialv

ar

v = 20 m/s r = 150 m μ = ?

Unbanked Curves

radial direction

ΣF = ma

fs = ma

μsFN = m(v2/r)

μsmg = mv2/r

μs = v2/gr

μs = (20m/s)2/((9.8 m/s2)(150m))

μs = 0.27

vertical direction

ΣF = ma

FN – mg = 0

FN = mg FN

fs

mg

vertical

radial

v = 20 m/s r = 150 m μ = ?

Unbanked Curves

33 A car goes around a curve of radius r at a constant speed v. Then it goes around the same curve at half of the original speed. What is the centripetal force on the car as it goes around the curve for the second time, compared to the first time?

A twice as bigB four times as bigC half as bigD onefourth as big

Banked Curves

Banking curves can help keep cars from skidding.

In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required.

Let's figure out what that speed is for a given angle and radius of curvature.

y

x θ

mgv

ar

Banked CurvesFront View

(the car heading towards you)Top View

We know the direction of our acceleration, so now we have to create axes.

Note that these axes will be different than for an Inclined Plane, because the car is not expected to slide down the plane.

Instead, it must have a horizontal acceleration to go in a horizontal circle.

a

Banked Curves

y

x θ

mg

vertical

radiala

Next, do the free body diagram.

Let's assume that no friction is necessary at the speed we are solving for.

Banked Curves

y

x θ

mg

vertical

radial

θ FN

mg

a

Next, decompose the forces that don't align with an axis, FN.

Banked Curves

y

x θ

vertical

radial

FN

mg

a

θFNcosθ

FNsinθ

Now let's solve for the velocity such that no friction is necessary to keep a car on the road while going around a curve of radius r and banking angle θ.

Banked Curves

y

x θ

vertical

radial

FN

mg

a

θFNcosθ

FNsinθ

radial direction

ΣF = ma

FNsinθ = ma

(mg/cosθ)(sinθ) = m(v2/r)

(g/cosθ)(sinθ) = (v2/r)

(gtanθ) = (v2/r)

v = (grtanθ)1/2

vertical direction

ΣF = ma

FNcosθ – mg = 0

FN = mg/cosθ

Practice Problem:Determine the velocity that a car should have while traveling around a frictionless curve with a radius 250 m is banked at an angle of 15°.

25.6 m/s

Conical Pendulum

A Conical Pendulum is a pendulum that sweeps out a circle, rather than just a back and forth path.

Since the pendulum bob is moving in a horizontal circle, we can study it as another example of uniform circular motion.

However, it will prove necessary to decompose the forces.

Draw a sketch of the problem, unless one is provided.

Then draw a free body diagram and indicate the direction of the acceleration.

θ

l

Conical Pendulum

θ

l

mg

T

a

Conical Pendulum

Then draw axes with one axis parallel to acceleration

θ

l

mg

T

a

Conical Pendulum

Then decompose forces so that all components lie on an axis...in this case, decompose T.

y

x

θ

l

mg

T

a

Conical Pendulum

Then solve for the acceleration of the bob, based on the angle θ, by applying Newton's Second Law along each axis.

y

x

θ

Tcosθ

Tsinθ

θ

l

mg

T

a

Conical Pendulum

y

x

θ

Tcosθ

Tsinθ

x direction

∑F = maTsinθ = ma

y direction

∑F = maTcosθ mg = 0Tcosθ = mg

Divide these two results

Tsin θ = maTcos θ = mg

tan θ = a/g

θ = tan1 (a/g) Or a = g tan θ

θ

l

mg

T

a

Conical Pendulum

y

x

θ

An alternative approach is to solve this as a vector equation using ΣF=ma. (This will work whenever only two forces are present.)

Just translate the original vectors (before decomposing them) to form a right triangle so that the sum of the two forces equals the new vector "ma".

θ

lmgT

Conical Pendulum

y

x

θ

Then, since

tan θ = opposite/adjacent

tan θ = ma/mg = a/g

tan θ = (v2/r)/g = v2/gr

v2 = grtan θ

the same result we found before

ma

θ

l

A 0.5 kg ball on a string is whirled in a circle with a radius of 0.75 m with a speed of 3 m/s.

Find the tension in the string.

θ

l

mg

T

a

Tx

Ty



xdirection ydirection

ΣF = ma ΣF = maTx = ma Ty mg = 0Tx = mv2/r Ty = mg

T2 = Tx2 + Ty2

T2 = (mv2/r)2 + (mg)2

T2 = m2(v4/r2 + g2)

T2 = (0.5kg)2((3m/s)4/(0.75m)2 + (9.8m/s2)2)

T2 = 60 N2

T = 7.7 N

θ

l



28.14 N

If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one.

Nonuniform Circular Motion

Uniform Circular Motion Power Point

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