Transcript Document

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

CENTRIPETAL FORCE.

It is a pull on the body and is directed toward the center of the circle. Force on car Force on passenger Tendency for passenger to go straight

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.

Remember Newton’s 1st Law?

What is the centripetal force?

-not a real force -feeling due to inertia

• • • • • • • • • • •

Circular Motion Vocabulary

r = radius m = mass v = velocity F c F T = centripetal force = tension force (sometimes written as T, not to be confused with the T for period) F f = friction force t T = time = period = “sec / rev” linear (tangential) velocity = 2πr/T “m / s” rotational (angular) velocity (ώ) = “rev / sec” = 1 / T a c = centripetal acceleration

Why does the ball stay in a circular path?

Centripetal:

(“c”) “center seeking” variables that exist when curved or circular path occurs . Top view

Linear (tangential) velocity

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

V

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

F

net-c

, a

c

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 → Linear velocity → T = time / # of rev (s/rev) v = 2πr / T (m/s)

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

Tension, mass, radius

Linear = Polar = Angular = Rotational 2πr 360º 2π rad 1 Rev Convert rotational speed to linear velocity: 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 …

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

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

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

Manipulate either Tension, m, r Control the other two variables Plot Tension, m, r vs. velocity

Procedure: (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!

Groups 3 masses, 3 tensions, 2 radius 0 Data table Time for 5 revolutions Period velocity F T ,r,m units Trial 1 Trial 2 Trial 3 s/1 rev m/s 0 0 0 0 0 1.) Find the Period: 2.) Find Linear Velocity:

Graph

F T vs v Control Hints r = .5 m m = .05 kg r vs v T = 2 N m = .05 kg m vs v T = 2 N r = .5 m Do not start with a (0,0) data point T (s/1rev) = time / # of rev.

v (m/s) = 2πr / T

Remember: Plot

velocity on the x-axis

Tension

F T

Radius

r v

Mass

m v v r m F T v 2 F T α v 2 r α v 2 v 2 1/v m 1/v 2 m α 1/v 2 For all Centripetal Forces:

F c = mv 2 /r

3 Important Equations

v = 2

p

r/T

Period (T) = time for 1 rev

Centripetal: (“c”) “center seeking”

F c and a c always act toward center

Velocity is tangent to the curve. Magnitude is constant, direction is not.

V ┴ F net-c , a c

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

Never put F c on the F.B.D.. Ie. F c = F f , F T , F N

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.

F N ΣF x -F T F T = ma = ma c c = m(-a c ) F T a c ΣF y F N = ma - F g = 0 = 0 F N = F g F g

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.

A F T ΣF x = ma c -(F T cosθ) = m(-a c ) F T cosθ = ma c a c F g ΣF y = ma F T sinθ - F g = 0 = 0 F T sinθ = F g

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.

F N ΣF x -F f F f = ma = ma c c = m(-a c ) F f a c F g ΣF y F N = ma - F g = 0 = 0 F N = F g

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.

F N a c ө ө F f F g ΣF x = ma c -(F N sinθ) – (F f cosө) = m(-a c ) F N sinθ + F f cosө = ma c ΣF y = ma = 0 F N cosθ - F f sinө - F g F N cosθ - F f sinө - F g = 0 = 0

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.

F f ΣF x -F N F N = ma = ma c c = m(-a c ) F N a c F g ΣF y F f = ma - F g = 0 = 0 F f = F g

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 #1

If 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 F T = 70.1 N

Problem #2

A 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.

F f = 8533.3 N μ = .726

Problem #3

A 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.

ө

Answer: v = 1.84 m/sec; T = 16.97 N

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

ө F N ө a c F g F f

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

201N 200N

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.

ΣF y = ma c F T + mg = mv²/r F T = mv²/r - mg F T = m ((v²/r) - g) ΣF y = ma c F T - mg = mv²/r F T = mv²/r + mg F T = m ((v²/r) + g)

a c F T mg a c F T mg

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

At this point, F T = 0, so… ΣF y = ma c F T + mg = mv²/r 0 + mg = mv²/r g = v 2 /r rg = v 2 or, v c =

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.

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?

7 m 49 m

60  y x 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 revolutions per minute?

 from the vertical (y), what is the rate of rotation in

Quiz

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.

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

20°

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?

Quiz #2

1.

A loop-the-loop rollercoaster has a radius of must have in order to stay on the track.

20 m

. Draw a FBD (at the top of the loop) showing all forces and calculate the minimum velocity the roller coaster v = __________ 2.

A satellite of

mass 1500 kg

planet of mass = 5

.7 x 10 22

is in low orbit around a

kg

and

radius 8.5 x 10

The altitude of the satellite is

4.7x 10 4 5 m

. gravitational attraction between the two? (

10 -11 N-m 2 /kg m

. What is the

G = 6.672 x

F g = __________