Uniform circular motion is the motion of an object in a circle with a constant speed.

What force or forces provides the circular motion in the following pictures?

What direction are these forces? Can it be more than one force? Could it be components of a force?

The force or forces needed to bend the normally straight path of a particle into a circular or curved path is called the

It is a pull on the body and is directed toward the center of the circle.

Tendency for passenger to go straight

Force on passenger

Force on car

CENTRIPETAL FORCE.

Without a centripetal force, an object in motion continues along a straight-line path.

With a centripetal force, an object in motion will be accelerated and change its direction.

What is the centripetal force?

Remember Newton’s 1st Law?

-not a real force-feeling due to

inertia

Circular Motion Vocabulary• r = radius• m= mass• v = velocity• Fc = centripetal force• FT = tension force

(sometimes written as T, not to be confused with the T for period)

• Ff = friction force• t = time• T = period = “sec / rev” • linear (tangential) velocity = 2πr/T “m / s”• rotational (angular) velocity (ώ) = “rev / sec” = 1 / T• ac = centripetal acceleration

Centripetal: (“c”) “center seeking” variables

that exist when curved or circular path occurs.

Why does the ball stay in a circular path?

What would happen to the ball if I cut the string?

V ┴ Fnet-c , ac

Linear (tangential) velocity is the velocity tangent to the curve at a point in time. (Newton’s 1st law)

Top view

An object is moving in a clockwise direction around a circle at constant speed.  

Which vector below represents the direction of the velocity vector when the object is located at point A, B, and C on the circle? the acceleration vector?the force vector?                                                                    

e.

Rotation → axis inside object

(Earth rotates around its axis)

Revolution → axis outside object

(Earth revolves around the Sun)

Rotational speed → rot. speed = # of rev / time (rev/s)

Period → T = time / # of rev (s/rev)

Linear velocity → v = 2πr / T (m/s)

What affects the speed of a ball tied to a string moving in a circular path?

Tension, mass, radius

Convert rotational speed to linear velocity:

Linear = Polar = Angular = Rotational

2πr 360º 2π rad 1 Rev

Example: The spinning ball (in the above picture) has a string radius of 0.5 m with the tube. There are 5 revolutions in 2.5 seconds (2 rev/s).

Convert to linear velocity …

Convert rotational speed of 5 rev / 2.5 s to linear speed in m/s. Remember T = 1 / rot. speed

Linear velocity: v = 2 π r / T

v = (2 π 0.5 / 0.5) = 6.28 m/s

Example: The spinning ball (in the picture at left) has a string radius of 0.5 m with the tube. The ball completes 5 revolutions in 2.5 seconds. (Rotational velocity)

r = 0.5 m

t = 2.5 seconds for 5 revolutions = T (period)

T = time/ # of rev = 2.5 s/ 5 rev = 0.5 s/rev

Manipulate either Tension, m, r

Control the other two variables

Plot Tension, m, r vs. velocityProcedure: (data table, plot and calculation on separate paper

• Find the time for 5 revolutions for each data point.

• Do 3 trials for each data point.

• Do 5 data points (15 total time trials)

• Convert rotational speed to linear velocity before plotting.

•Determine the relationship of the graph – explain!

Purpose: Find the relationship between the _________variable and the velocity of the mass.

Data tableFT,r,m Time for 5 revolutions Period velocity

units Trial 1 Trial 2 Trial 3 s/1 rev m/s

0 0 0 0 0 0

Control Hints

FT vs v

r vs v

m vs v

Graphr = .5 m

m = .05 kg

T = 2 N

m = .05 kg

T = 2 N

r = .5 mDo not start with a (0,0) data point

1.) Find the Period: T (s/1rev) = time / # of rev.

2.) Find Linear Velocity: v (m/s) = 2πr / T

Remember: Plot velocity on the x-axis

Groups 3 masses, 3 tensions, 2 radius

Tension

Radius

Mass

FTFT

v

m

v

v

r

m

v2

v2

1/v

r

m

1/v2

FT α v2

r α v2

m α 1/v2

For all Centripetal Forces:

Fc = mv2/r

3 Important Equations

v = 2r/T

Period (T) = time for 1 rev

•Centripetal: (“c”) “center seeking”

•Fc and ac always act toward center

•Velocity is tangent to the curve.

Magnitude is constant, direction is not.

•V ┴ Fnet-c, ac

•Radius is the ┴ distance from the axis of rotation to the object

•Never put Fc on the F.B.D.. Ie. Fc = Ff , FT , FN

•Projectile Motion is not true Circular motion, Why?In circular motion the force is always ┴ to the velocity; the

magnitude of velocity is not effected by the force

1. What does centripetal mean?

A ball is attached to a string and whirled around at a constant speed in a circle with a radius 3 meters. (Questions 2-6)

2. What is the direction of the net force on the ball?

3. What is the direction of the accel. of the ball? Is it constant?

4. What is the direction of the velocity of the ball? Is it constant?

5. What are the 3 important equations? Does rotational speed depend upon the radius? Does linear speed depend upon the radius?

6. If it takes 50 seconds for the ball to make 20 revolutions around a 3 m radius, calculate the rotational speed and linear velocity of the ball.

7. What were the relationships we found in the lab?

4. Tied to a post and moving in a circle at constant speed on a frictionless horizontal surface. Coming straight out of the paper.

FT

Fg

ac

ΣFy = maFN - Fg = 0FN = Fg

ΣFx = mac

-FT = m(-ac)FT = mac

= 0

FN

5. Tied to point A by a string. Moving in a horizontal circle at constant speed. Not resting on a solid surface. No Friction. Coming straight out of paper.

FT

Fg

ac

ΣFy = maFTsinθ - Fg = 0FTsinθ = Fg

ΣFx = mac

-(FTcosθ) = m(-ac)FTcosθ = mac

= 0

A

10. Riding on a horizontal disk that is rotating at constant speed about its vertical axis. Friction prevents rock from sliding. Rock is moving straight out of the paper.

FN

Fg

ac

ΣFy = maFN - Fg = 0FN = Fg

Ff

ΣFx = mac

-Ff = m(-ac)Ff = mac

= 0

11. Resting against a wall with friction a cone is rotating about its vertical axis at a constant speed. Not accelerating vertically. Moving straight

out of the paper.FN

Fg

ac

ΣFy = maFNcosθ - Ffsinө - Fg = 0FNcosθ - Ffsinө - Fg = 0

ΣFx = mac

-(FNsinθ) – (Ffcosө) = m(-ac)

FNsinθ + Ffcosө = mac

= 0

ө

Ff

ө

12. Stuck by friction against the inside wall of a drum rotating about its vertical axis at constant speed. Rock is moving straight out of the paper.

Ff

Fg

ac

ΣFy = maFf - Fg = 0Ff = Fg

FN

ΣFx = mac

-FN = m(-ac)FN = mac

= 0

A ball held by a string is coasting around in a large horizontal circle. The string is then pulled in so the ball coasts in a smaller circle. When it is coasting in the smaller circle its speed is … (Assume tension and mass stay constant)

a) greater

b) less

c) Unchanged

Explain.

Problem #1If the radius of a circle is 1.5 m and it takes 1.3 seconds for a mass to swing around it (1 rev).

a) What is the speed of the mass? b) Find the tension if the mass is 2 kg.

s = 7.25 m/s

FT = 70.1 N

Problem #2A 1200 kg car traveling at 8 m/s is turning a corner with a 9 m radius.

a) How large a force is needed to keep the car on the road?

b) b) Find the coefficient of friction.

Ff = 8533.3 Nμ = .726

Problem #3A car travels around a circular flat track with a speed of 20 m/s. The coefficient of friction between the tires and the road is 0.25. Calculate the minimum radius needed to keep the car on the track.

r = 163.27 m

What speed must a 1.5 kg pendulum bob swing in the circular path of the

accompanying figure if the supporting cord is 1.2 m long and is 30? Also find

the tension in the cord.

Answer:

v = 1.84 m/sec;

T = 16.97 N

ө

The Gunslinger is modeled after the famed Flying Dutchmen rides of carnival midways. Guests ride in individual chairs suspended by tempered steel chains. The arms tilt to a 25 angle. As a safety

engineer for the Six Flags of America Corporation, you are asked to determine the maximum

allowable rotation rate for the Gunslinger if the breaking strength of the steel chains are 1000 N.

Other data:

length of chain and swing: 4.5 m

distance from center of rotation to chain attachment: 6.7 m

1. Draw a FBD of a rider and the swing 2nd Law Equations.

2. What is the source of the centripetal force acting on a rider and the swing?

3. Which will ride higher: an empty swing or one with someone in it? Explain.

4. Determine the maximum allowable rotation rate.

Banked Road problem

FN

FfFg

acө

ө

201N

200N

Prove that the scale reading is greater at the poles than at the equator.

Bill the Cat, tied to a rope, is twirled around in a vertical circle. Draw the free-body diagram for Bill

in the positions shown. Then sum the X and Y forces.

ΣFy = mac

FT + mg = mv²/r

FT = mv²/r - mg

FT = m ((v²/r) - g)

ΣFy = mac

FT - mg = mv²/r

FT = mv²/r + mg

FT = m ((v²/r) + g)

FT

mg

ac

mgFT

ac

Minimum velocity needed for an object to continue moving in a vertical circle. Any less

velocity and the object will fall.

At this point, FT = 0, so…

ΣFy = mac

FT + mg = mv²/r

0 + mg = mv²/r

g = v2/r

rg = v2 or, vc = rg

Suppose a car moves at a constant speed along a mountain road. At what places does it exert the greatest and least forces on the road?

a) the top of the hill

b) at the dip between two hills

c) on a level stretch near the bottom of the hill

Explain each case with a free body diagram and sum the forces.

1) Draw a FBD of the Greezed Lightning at the top of the loop.

2) Write the 2nd Law Equation for the FBD.

3) List the assumptions made and calculate the minimum velocity of the Greezed Lightning given the following picture.

7 m

49 m

4) What if you had designed a Greezed Lightning type rollercoaster and the coaster was moving too slow, how could you change your design to correct the problem?

An automatic tumble dryer has a 0.65 m diameter basket that rotates about a horizontal axis (x). As the basket turns, the clothes fall away from the basket’s edge and tumble over. If the clothes fall away from the basket at a point 60 from the vertical (y), what is the rate of rotation in revolutions per minute?

y

x

60

Quiz

2. A ball is swung in a horizontal circle. The 1.5 m long string makes a 20° to the horizontal.

a. Draw all forces on the picture, including direction of acceleration.

b. Write the summation and net force equations (x & y)

c. Determine the velocity of the ball?

1. A race car travels around a flat, circular track with a radius of 180 m. ….The coefficient of friction between the tires and the pavement is 1.5.

a. Draw all forces on the car, including direction of acceleration.

b. Write the summation and net force equations (x & y).

c. Calculate the maximum velocity the car can go and stay on the track.

20°

Quiz #2

1. A loop-the-loop rollercoaster has a radius of 20 m. Draw a FBD (at the top of the loop) showing all forces and calculate the minimum velocity the roller coaster must have in order to stay on the track.

v = __________

2. A satellite of mass 1500 kg is in low orbit around a planet of mass = 5.7 x 1022 kg and radius 8.5 x 105 m. The altitude of the satellite is 4.7x 104 m. What is the gravitational attraction between the two? (G = 6.672 x 10-11 N-m2/kg

Fg = __________