Circular MotionObjectives: Use vectors to pictorially analyze problems for circular and rotary motion.

Solve circular motion word problems.

Interpret the relationship between the force on an object and the velocity of the object for circular motion through the analysis of experimental data.

Circular Motion

Suppose you drive a go cart in a perfect circle at a constant speed. Even though your speed isn’t changing, you are accelerating. This is because acceleration is the rate of change of velocity (not speed), and your velocity is changing because your direction is changing! Remember, a velocity vector is always tangent to the path of motion.

v v

v

Centripetal AccelerationFormula for centripetal acceleration

in uniform circular motion

Acceleration, a = (vf - v0) / t. We are subtracting vectors here, not speeds, otherwise a would be zero. (v0 and vf have the same magnitudes.) The smaller t is, the smaller will be, and the more the blue sector will approximate a triangle. The blue “triangle” has sides r, r, and v t (from d = v t ). The vector triangle has sides v, v, and | vf - v0 |. The two triangles are similar (side-angle-side similarity).

v0 = v

r

r

vf = v

vf

v0

vf - v0

r

r

v t

continued on next slide

Centripetal Acceleration (cont.)

v0 = v

r

r

vf = vBy similar triangles,

v

v

| v f - v0 |

r

r

v t

v r

=| vf - v0 |

v t

So, multiplying both sides above by v, we have

| vf - v0 |

t=

rv

2ac =

Unit check:(m/s)2

m=

m2 / s2

m

m

s2=

Tangential vs. Centripetal Acceleration

10 m/s

15 m/s

So how do we calculate the centripetal acceleration ? ? ? Stay tuned!

18 m/s

start

finish

Suppose now you drive your go cart faster and faster in a circle. Now your velocity vector changes in both magnitude and direction. If you go from start to finish in 4 s, your average tangential acceleration is:

at = (18 m/s - 10 m/s) / 4 s = 2 m/s2

So you’re speeding up at a rate of 2 m/s per second. This is the rate at which your velocity changes tangentially. But what about the rate at which your velocity changes radially, due to its changing direction? This is your centripetal (or radial) acceleration.

Centripetal acceleration vector always points toward center of circle.

vac at

vac

atmoving counterclockwise;

speeding upmoving counterclockwise;

slowing down

“Centripetal” means “center-seeking.” The magnitude of ac depends on both v and r.

However, regardless of speed or tangential acceleration, ac always points toward the

center. That is, ac is always radial (along the radius).

Resultant Acceleration

ac at

a

The overall acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. That is,

a = ac + at

This is true regardless of the direction of motion. It holds true even when an object is not moving in a perfect circle. Note: The equation above does not include v. Vectors of different quantities cannot be added!

moving counterclockwise while

speeding up or moving clockwise while slowing down

v

ac

ac

Non-circular paths

ac

R2

R1

P1 v

P2

Here we have an object moving along the brown path at a constant speed (at = 0). ac

changes, though, since the radius of curvature changes. At P1 the path is approximated by the large green circle, at P2 by the smaller orange one. The smaller r is, the bigger ac is.

r

v 2

ac =

Friction and ac

ac 20 m/s

60 m

You’re cruising at a constant 20 m/s on a winding highway. The radius of curvature where you are is 60 m. Your centripetal acceleration is:

ac = (20 m/s)2 / (60 m) = 6.67 m/s2

The force that causes this acceleration is friction, which is why it’s hard to turn on ice. Friction, in this case, is the “centripetal force.” The sharper the turn or the greater your speed, the greater the frictional force must be.

overhead view

continued on next slide

Friction and ac (cont.)

ac v

r

overhead view

f

Since you’re not speeding up, f is the net force, so Fnet

= f = s N = s mg = ma. We use s because you’re not sliding (or even moving) radially. Thus,

s mg = ma.

Mass cancels out, showing that your centripetal acceleration doesn’t depend on how heavy your vehicle is. Solving for a we have

a = s g.

In the diagram N is pointing out of the slide. Also, a = ac

in this case since at = 0.

m

Centripetal Force, Fc

From F = ma, we get Fc = mac = mv2 / r.

Fc =mv

2

r

If a body is turning, look at all forces acting on it, and find the net force. The component of the net force that acts toward the center of curvature (perpendicular to the body’s motion) is the centripetal force. The component that acts parallel to its motion (forward or backwards) is the tangential component of the net force.

Forces that can provide a centripetal force

• Friction, as in the turning car example• Tension, as in a rock whirling around while attached to a string,

or the tension in the chains on a swing at the park.*• Normal Force, as in a “round-up ride” at an amusement park

(that spins & the floor drops out), or the component of normal force on a car on a banked track that acts toward the center.*

• Gravity: The force of gravity between the Earth and sun keeps the Earth moving in a nearly circular orbit.

• Any force directed toward your center of curvature, such as an applied force.

* Picture on upcoming slides

Loop-the-loop in a Plane

mg

mg Ntop

A plane flies in a vertical circle, so it’s upside down at the high point. Its speed is constant, but because of its nonlinear motion, the pilot must experience centripetal acceleration. This ac is provided by a combination of her weight and her normal force. mg is constant; N is not. N is the force her pilot’s chair exerts on her body.

v

v

continued on next slide

Nbot

Loop-the-loop (cont.)

mg

Nbot

mg Ntop

v

vTop: Normal force and weight team up to provide centripetal force: Ntop + mg = mv2 / r. If the pilot were sitting on a scale, it would say she’s very light.

Bottom: Weight works against normal force, so N must be bigger down here to provide the same centripetal force: Nbot - mg = mv2 / r. (Fc has a constant magnitude since m, v, and r are constants.) Here a scale would say that the pilot is very heavy.

continued on next slide

r

Loop-the-loop (cont.)

We’ve been discussing the pilot, but what force causes the plane to turn?

Answer:

The normal force (force on pilot due to seat) changes throughout the loop. This case is similar to the simple pendulum (the only difference being that speed is constant here). Part of the weight opposes N, and the net radial force is the centripetal force:

N - mg cos = mac = mv2 / r

mg

N

r

mg cos mg sin

The air provides the centripetal force on plane

T

m

g co

sSimple Pendulum

m

mg

mg sin

Two forces act on a swinging pendulum, tension and weight. Tension acts radially. We break the weight vector into a radial component (green) and tangential one (violet). Blue is bigger than green, otherwise there would be no net centripetal force, and the mass couldn’t turn.

Fc = T - mg cos = mac = mv2 / L.

Ft = mg sin = mat Ft is the force that speeds the mass up or slows it down. The weight is a constant, but since changes, so do the weight’s components. mg cos is greatest when the mass is at its low point. This is also where it’s moving its fastest. For these two reasons T is the greatest when the mass is at the low point.

string of length L

• A pendulum’s period is the same for big arcs as it is for little arcs, so long as the angle through which it swings isn’t real large. (The average speed, therefore, is greater when the arc is bigger, because it must cover a bigger distance in the same time.)

• The period is independent of the mass.• The period depends only on the pendulum’s length.

• The period = T = 2 (proven in advanced physics)

• This formula gives us a way to measure the acceleration due to gravity (which varies slightly with location) by measuring the period and length of a pendulum.

• Don’t confuse the symbol T, which is used for both period and tension.

Facts about the Simple Pendulum

g

L

Conical Pendulum

T

mg

ac

T

mg

ac

m m

Like a tether ball, the mass hangs from a rope and sweeps out a circular path, and

the rope a cone. is a constant. The vertical component of T balances mg. The horiz. comp. of T is the centripetal

force.

vv

continued on next slide

Conical Pendulum (cont.)

T

mg

T sinm

T cos

All vectors shown are forces. v points into or out of the slide, depending on the direction in which the mass moves. ac is parallel to T sin, which serves as the centripetal force.

T cos = mg

T sin = mv2 / r

Dividing equations: tan = v2 / rg

See it in action

Gravity

Newton’s Law of Gravitation

Kepler’s Laws of Planetary Motion

Gravitational Fields

Newton’s Law of Gravitation

m1

m2r

There is a force of gravity between any pair of objects anywhere. The force is proportional to each mass and inversely proportional to the square of the distance between the two objects. Its equation is:

FG = G m1 m2

r 2

The constant of proportionality is G, the universal gravitation constant. G = 6.67 · 10-11 N·m2 / kg2. Note how the units of G all cancel out except for the Newtons, which is the unit needed on the left side of the equation.

Gravity Example

FG = G m1 m2

r 2

How hard do two planets pull on each other if their masses are 1.23 1026 kg and 5.21 1022 kg and they 230 million kilometers apart?

This is the force each planet exerts on the other. Note the denominator is has a factor of 103 to convert to meters and a factor of 106 to account for the million. It doesn’t matter which way or how fast the planets are moving.

(6.67 · 10-11 N·m2 / kg2) (1.23 · 1026 kg) (5.21 · 1022 kg) =(230 · 103 · 106 m)

2

= 8.08 · 1015 N

3rd Law: Action-ReactionIn the last example the force on each planet is the same. This is due to to Newton’s third law of motion: the force on Planet 1 due to Planet 2 is just as strong but in the opposite direction as the force on Planet 2 due to Planet 1. The effects of these forces are not the same, however, since the planets have different masses.

For the big planet: a = (8.08 · 1015 N) / (1.23 · 1026 kg) = 6.57 · 10-11 m/s2.

For the little planet: a = (8.08 · 1015 N) / (5.21 · 1022 kg) = 1.55 · 10-7 m/s2.

5.21 · 1022 kg

1.23 · 1026 kg8.08 · 1015 N 8.08 · 1015 N

Inverse Square Law

FG = G m1 m2

r 2

The law of gravitation is called an inverse square law because the magnitude of the force is inversely proportional to the square of the separation. If the masses are moved twice as far apart, the force of gravity between is cut by a factor of four. Triple the separation and the force is nine times weaker.

What if each mass and the separation were all quadrupled?

answer: no change in the force

Calculating the Gravitational ConstantIn 1798 Sir Henry Cavendish suspended a rod with two small masses (red) from a thin wire. Two larger mass (green) attract the small masses and cause the wire to twist slightly, since each force of attraction produces a torque in the same direction. By varying the masses and measuring the separations and the amount of twist, Cavendish was the first to calculate G.

Since G is only 6.67 · 10-11 N·m2 / kg2, the measurements had to be very precise.

Calculating the mass of the EarthKnowing G, we can now actually calculate the mass of the Earth. All we do is write the weight of any object in two different ways and equate them. Its weight is the force of gravity between it and the Earth, which is FG in the equation below. ME is the mass of the Earth, RE is the radius of the Earth, and m is the mass of the object. The object’s weight can also be written as mg.

=FG = G m1 m2

r 2

G ME m RE

2 = m g

The m’s cancel in the last equation. g can be measured experimentally; Cavendish determined G’s value; and RE can be calculated at 6.37 · 106 m (see next slide). ME is the only unknown. Solving for

ME we have: g RE

2

GME = = 5.98 · 1024 kg

Rotational Motion

Arc length: s = r

r

s

If is in radians, then the arc length, s, is times r. This follows directly from the definition of a radian. One radian is the angle made when the radius of a circle is wrapped along the circle.

1 radian

r

r

When the arc length is as long as the radius, the angle subtended is one radian. (A radian is really dimensionless, since it’s found by dividing a length by a length.)

Angular Speed, Linear speed is how fast you move, measured as distance per unit time. Angular speed is how fast you turn, measured as an angle per unit time. The symbol for angular speed is the small Greek letter omega, , which looks like a curvy “w”. Units for angular speed include: degrees per second; radians per second; and rpm (revolutions per minute).

3 m110

A

B

Suppose an object moves steadily from A to B along the circle in 5 s. Then = 110 / 5 s = 22 / s. The distance it covers is (110 / 360)(2)(3) = 5.7596 m. So its linear speed is v = (5.7596 m) / (5 s) = 1.1519 m / s.

v = r

M

C

C´The Three Stooges go to the park. Moe and Larry are on a merry-go-round ride of radius r that Larry is pushing counterclockwise, running at a speed v. In a time t, Moe goes from M to M´ and Curly goes from C to C´. Moe is twice as far from the center as Curly. The distance Moe travels is r , where is in radians. Curly’s distance is ½ r . Both stooges sweep out the same angle in the same time, so each has the same angular speed. However, since Moe travels twice as far, his linear speed in twice as great.

M´

v

v / 2

s = r

st

r t= =

t

r

v = r

Ferris Wheel Problem

Schmedrick is working as a miniature ferris wheel operator (radius 2.1 m). He gets a little overzealous and cranks it up to 75 rpm. His little brother Poindexter flies out at point P, when he is 35 from the low point. At the low point the wheel is 1 m off the ground. A 1.5 m high wall is 27 m from the low point of the wheel. Does Poindexter clear the wall?

P

Strategy outlined on next slide

Ferris Wheel Problem-Solving Strategy

1. Based on his angular speed and the radius, calculate Poindexter’s linear speed.

2. Break his launch velocity down into vertical and horizontal components.

3. Use trig to find the height of his launch.

4. Use trig to find the horizontal distance from the launch site to the wall.

5. Calculate the time it takes him to go that far horizontally.

6. Calculate height at that time.

7. Draw your conclusion.

16.4934 m/s

vx = 13.5106 m/s vy0 = 9.4602 m/s

1.3798 m

25.7955 m

1.9093 s

1.5798 m

He just makes it by about 8 cm !