Newton’s Second Law:Newton s Second Law:

Uniform Circular Motion

Readings: Chapter 51

Readings: Chapter 5

Kinematics in Two Dimensions: Uniform Circular Motion: Motion with Constant Speed

The speed (magnitude of velocity) is the same

Period:

2 rπ2 rTvπ

=

θ

The position of the particle is characterized by angle (or by arc length)

Arc length:

where angle is in RADIANS

s rθ=

θ2

g0

03601 57.32

radπ

= =

Example:

What is the arc length for if0 0 045 90 10θ = 1r m=What is the arc length for if

0 00

245 45 0.78360 4

rad rad radπ π= = =

45 ,90 ,10θ = 1r m=

360 40 0

0

290 90 1.57360 2

rad rad radπ π= = =

0 00

210 10 0.174360 18

rad rad radπ π= = =

Arc length: s rθ=rπ

45 0.784rs mπ= =

90 1.572rs mπ= =

10 0.17418rs mπ= =

3

90 2

Kinematics in Two Dimensions: Uniform Circular Motion: Motion with Constant Speed

The speed (magnitude of velocity) is the same s vt=tθ ω= tθ ω

vr

ω = 2 rTvπ

=

Then

2π2Tπω =

4

Uniform Circular Motion s vt= 2Tπω =

tθ ω=

vr

ω =

Then

cos cos( )x r r tθ ω= = ( )

sin sin( )y r r tθ ω= =

2 2 2 2 2 2 2

5

2 2 2 2 2 2 2sin ( ) cos ( )x y r t r t rω ω+ = + =

v∆

Uniform Circular Motion: Acceleration

r∆

0t

vat ∆ →

∆=∆ 0t

rvt ∆ →

∆=∆

1 1r v t∆ = ∆

2 2r v t∆ = ∆2 2r v t∆ ∆2 1 2 1

2( )v v r r

at t− ∆ − ∆

= =∆ ∆( )t t∆ ∆

Direction of acceleration is the same as direction of vector (toward the center)

→

2 1CB r r→

= ∆ − ∆The magnitude of acceleration:

2( 2 )CB φ θ∆ ∆ ∆ ∆ ∆6

2

2 2 2 2 2

( 2 )( ) ( ) ( ) ( ) ( )CB r r r v t t va vt t t t t r

φ π α θ ωω

∆ ∆ − ∆ ∆ ∆= = = = = = =

∆ ∆ ∆ ∆ ∆

Uniform Circular Motion: Acceleration vr

ω =

centripetal acceleration

The magnitude of acceleration:

22va r v

rω ω= = =

7

Uniform Circular Motion: Dynamics vr

ω =

The magnitude of acceleration:netF

22va r v

rω ω= = =

F

netF

Direction of acceleration: toward the center

netF

The second Newton’s law:

8netF ma=

Uniform Circular Motion: Dynamics2v

The magnitude of the net force

2va r vr

ω ω= = =

The magnitude of the net force2

netvF mr

=

Direction of the net force: toward the center

r

netF n T w ma= + + =

0n w− =2vT ma m= =

9

T ma mr

Uniform Circular Motion: Dynamics

tFnetFn

fnetF n f w ma= + + =

f0n w− =2vf ma m= =

wf ma m

r

maxf n w mgµ µ µ= = =

then2vm mgµ≤m mgr

µ≤

v rgµ≤

10

gµ

Fictitious Forces

The force that seems to push the object outside of the circle is called centrifugal force

The direction of centrifugal force is opposite to the direction of acceleration

The centrifugal force exists only inside the systems which is moving in a circle (inside the car)

The centrifugal force is not a REAL FORCE

The magnitude of centrifugal force is

2vF m=11

cfF m r=

Fictitious Forces: generalized second Newton’s law

In the reference frame of the car (inside the car) there are no motion, so there are no acceleration,there are no motion, so there are no acceleration, but THERE IS CENTRIFUGAL FORCE

nn

f

0net cfF n f w F= + + + =

cfF

w

0n w− =

0f F− = w0cff F− =

2

fvF m=

12

cfF m r