Chapter 5Uniform Circular Motion

ac=v2/r

Uniform circular motion:Motion in a circular path with constant speed

1) Speed and period

• Period, T: time for one revolution• Speed is related to period:

– Path for one revolution: s = 2πr– Speed: v = s/T = 2πr/T

rs v

2) Centripetal Acceleration

• Directed toward centre (always changing)• Magnitude:

rs

ac

ac =v2

r

v

Two cars, driving at the same constant speed v, are traveling through turns that have different r, as shown in the drawing. Which statement is true about the magnitude of the centripetal acceleration ac of each car?

(a) ac of the car at A is greater than that of the car at B, since r is greater at A.

(b) ac of the car at A is smaller than that of the car at B, since r is smaller at A.

(c) ac of each car is the same, since both cars are moving at v.

(d) ac of the car at A is greater than that of the car at B, since r is smaller at A.

3) Centripetal force

• Toward centre (parallel to acceleration) since F = ma (always changing)

• Magnitude:

rs

acFc

Fc = mac =mv2

r

v

• Centripetal force is a real force that produces circular motion in an inertial frame

gravity Tension Friction, Normal

• Centrifugal force is a virtual force perceived in a rotating (non-intertial frame)

Imagine you are swinging a bucket by the handle around in a circle that is nearly level with the ground (a horizontal circle). Now imagine there's a ball in the bucket. What keeps the ball moving in a circular path?

A. contact force of the bucket on the ball

B. contact force of the ball on the bucket

C. gravitational force on the ball

D. the centrifugal force

C&J 5.59 A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 28 m/s, the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 150 m), the block swings toward the outside of the curve. Then the string makes an angle θ with the vertical. Find θ.

4) Driving around circular curvesa) Unbanked curve

- static friction provides Fc

vmax = µsgr

!fs

!FN

m!g

Fc = fs ≤ µsFN = µsmg

r

Fc = mv2 / r

b) Banked curve, no friction

tanθ = v2

rgx :FN sinθ = mv2 / r

y :FN cosθ = mg

x

y

c) Banked curve with friction

Recall ideal angle:

tanθ =v2

rg

i) If v2 > gr tan θ, friction prevents sliding up

fs(i)

ii) If v2 < gr tan θ, friction prevents sliding down

fs(ii) !F =!FN + m!g +

!fs

Net force:

Example: Find maximum velocity for µ = 0.80, θ = 47º, and r = 60 m

mg

FN

fs

θ

Fc θ !Fc =

!FN + m!g +

!fsx

y

fs(i)

fs(ii)

y-motion

0 = FN cosθ − mg − fs sinθWhen v = vmax , fs = µFN so

FN =mg

cosθ − µ sinθ

x-motion

Fc = FN sinθ + fs cosθ

When v = vmax , fs = µFN , and using Fc=mv2

r,

FN =mvmax

2

r(sinθ + µ cosθ )

Eliminate FN , and solve for vmax :

vmax2 = gr sinθ + µ cosθ

cosθ − µ sinθ⎛⎝⎜

⎞⎠⎟

vmax = 88m/s = 317km/h

0 = FN cosθ − mg − fs sinθWhen v = vmax , fs = µFN so

FN =mg

cosθ − µ sinθ

Fc = FN sinθ + fs cosθ

When v = vmax , fs = µFN , and using Fc=mv2

r,

FN =mvmax

2

r(sinθ + µ cosθ )

5) Satellites in circular orbitsa) Speed and radius

F = G mME

r2= mv2 / r

v = GME

r

• Radius of orbit determines speed (independent of mass)

• Equal acceleration of all local objects

--> weightlessness

Law of gravity circular motion

m

ME

v = 2πrT

= GME

r

T = 2πr32

GME

speed and period orbit

(b) Period and radius

c) Radius of synchronous orbits

•  satellite is stationary above earth’s surface

• Conditions: • T = 1 sidereal day• above equator

r32 =

T GME

2π

Using T = 24h − 4 min = 86160 s

gives r = 4.22 ×107 m = 6.61RE

It is the year 2094; and people are designing a new space station that will be placed in a circular orbit around the Sun. The orbital period of the station will be 6.0 years. Determine the ratio of the station’s orbital radius about the Sun to that of the Earth’s orbital radius about the Sun. Assume that the Earth’s orbit about the Sun is circular.

a) 2.4 d) 5.2 b) 3.3 e) 6.0 c) 4.0

C&J 5.52 A child is twirling a 0.0120-kg plastic ball on a string in a horizontal circle whose radius is 0.100 m. The ball travels once around the circle in 0.500 s.

(a) Determine the centripetal force acting on the ball.

(b) If the speed is doubled, does the centripetal force double? If not, by what factor does the centripetal force increase?

C&J 5.11 A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container.

Suppose the centripetal acceleration of the sample is 6.25 × 103 times as large as the acceleration due to gravity.

How many revolutions per minute is the sample making, if it is located at a radius of 5.00 cm from the axis of rotation?

October 2008 Midterm exam  Q18 A 1200.0 kg car, traveling at 30.0 m/s, travels safely (without skidding) around a frictionless banked curve of radius 200.0 m. What is the magnitude of the normal force on this car?

6) Centrifugal force and artificial gravitya) Artificial gravity

Fc = FN (inside surface)

e.g. if r = 1700 m, and Fc = mg, then mg = mv2/r, giving

v = gr = 130 m/s = 468 km/h

Centripetal force acts toward the centre (up, like the normal force on earth) but gravity acts down. What has taken the place of gravity, from the perspective of the cylinder-dwellers?

Centrifugal force…does not exist (in inertial frames)is a fictitious, pseudo, virtual forceis an inertial force

An episode of Quirks & Quarks on CBC radio illustrates the confusion that can exist about artificial gravity.

The question was: “What happens when you jump inside a rotating cylinder like the one on the movie 2001: A Space Odysey”

“So, what? He’ll go straight across and his head will smash into the other side?”

“That’s right.” No. That’s wrong.

The simulated gravity does not require continuous contact with the surface. From your perspective (the jumper’s), you will be pulled back to the surface just as you would in a gravitational field, apart from a small deviation depending on the radius of the cylinder. To understand this, we will consider a simpler inertial force first.

Inertial frame

m a0

b) Inertial force -- Apparent force resulting from an accelerating reference frame.

e.g. An accelerating spaceship (far from planets)

FN = ma0

a0FN

Inside the ship (zero acceleration)

F = ma∑ = 0?

To an observer inside the ship an inertial force equal to ma0 is acting toward the floor.

FN −ma0 = 0ma0

F = ma∑ FN −ma0 = 0

(A dropped ball falls.)

c) Centrifugal force: the inertial force in a rotating reference frame

In an inertial frame, only the normal force and gravity act to give centripetal force which produces centripetal acceleration (no such thing as centrifugal force)

Consider a cup of water swung around in a vertical circle. What keeps the water in the cup when it is upside-down?

FN

mgAnswer: Inertia. The cup pulls the water down faster than its natural falling rate.

FN + mg = mv2

r

Gravity plus the normal force provide centripetal force that produces circular motion:

What keeps water in the cup from the perspective of the cup-dweller?

Answer: Invent centrifugal force to balance real forces and ensure zero acceleration. So

FN +mg −mv2

r= 0

F∑ = 0

mv2/r

FN

mg

What happens if a ball is dropped?

If the force holding it Fh is removed, it should fall radially (down to the cup-dweller), but in an inertial frame it moves tangentially.

FN

mg

How can both be true?

Fh

mv2/r

In the rotating reference frame it accelerates radially outward (down to the observer), apart from a small curvature which decreases for larger cylinder radii.

An object released moves on the tangent,according to Newton’s first law.

an object launched radially (jumping) moves with constant velocity at a small angle to the tangent

moves up and down in the rotating frame (except for small displacement due to finite radius)

An accelerating reference frame can duplicate the effect of gravity (not only when in contact).

31

inertial frame rotating frame

7) Vertical circular motion

• Not usually uniform motion since the speed is changing• Net force not always toward the centre• Component of acceleration toward the centre (the

centripetal component) is still v2/r, soFc =

mv2

r

FN 4 =mv4

2

r

FN 3 + mg =mv3

2

r

FN1 − mg =mv1

2

r

FN 2 =mv2

2

r

FN 3 + mg =mv3

2

r

Minimum speed at top:FN 3 = 0

mg = mv32

rv3 = rg

For r = 6 m, v = 7.6 m/s or 27 km/h

In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle (radius r = 0.32 m) , as the drawing shows. The appliance is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches an angle of above the horizontal, it loses contact with the wall of the cylinder and falls onto the clothes below.

How many revolutions per second should the cylinder make in order that the clothes lose contact with the wall when theta = 70.0º?

r = 0.32 mθ = 70º