General Physics

Today:

1. Circular Motion / Rotation

Rotational Kinematics

Definition of position, velocity and acceleration for rotational motion.

Rotational kinematics Linear kinematics

Angular Position θ (theta) x

Angular Velocity ω (omega) v

Angular Acceleration α (alpha) a

Angular Position Θ

Θ=0 : reference line

Θ > 0 : counterclockwise

Θ < 0 clockwise

Units: degree °

revolution rev (1rev = 360 °)

radian (most convenient unit)

A radian is the angle for which the arc length s on a circle is equal to the radius r

s=r Θ (s=r for 1 rad) and therefore Θ=s/r [m/m=rad]

1rev=360°= 2 π rad or 1rad ≡ 180°/π

1rad = 57.3°

Revolution, Period , Frequency and Angular Velocity

ΔΘ = Θf - Θi (change of angular position)Revolution rev = 360°period T (time taken for one revolution) [s] frequency f (how many revolutions per second) [s-1] = [Hz] are related by: 1 = T f

Angular velocity ω = ΔΘ/Δt (average; instantaneous for lim Δt → 0)

Average velocity for one period Tπω 2

=

][s

rad

Sign of the angular velocityω > 0 : counterclockwiseΔΘ > 0 (change of angular position)

ω < 0 clockwiseΔΘ < 0 (change of angular position)

Units: rad/s = 1/s

avav timetotaldistance total ωθθ r

tr

trv =⎟

⎠⎞

⎜⎝⎛ΔΔ

=ΔΔ

==

An object moves along a circular path of radius r; what is its average speed?

Linear Kinematics vs Rotational Kinematics

The block falls down with a constant acceleration

The pulley turns right (clockwise) with a constant angular acceleration

tavvtva i +=ΔΔ= f or /

tt i αωωωα +=ΔΔ= f or /

Velocity of box is equal the tangential velocity of the pulley

Here, Δv≠0. The direction of v is changing.

If Δv≠0, then a≠0. The net force cannot be zero.

Consider an object moving in a circular path of radius r at constant speed.

x

y

v

v

v

v

Conclusion: to move in a circular path, an object must have a nonzero net force acting on it.

Centripetal Acceleration

The velocity of a particle is tangent to its path.

For an object moving in uniform circular motion, the acceleration is radially inward.

vrrvar ωω === 2

2

The magnitude of the radial acceleration is:

Centripetal Acceleration of an Rotating Object

acp= v2/r = (rω)2/r

acp = r ω2 [m/s2]

Tangential acceleration of an accelerating object

Δv = r Δω

at = Δv / Δt = r Δω / Δt = r α [m/s2]

tan

1

22

tanaa

aaa

cp

cpt

−=

+=

φ

φ Angle

onacceleratiTotal

Centripetal Force

According to Newton's second law of motion, an accelerating object must be acted upon by an unbalanced force. This unbalanced force is in the same direction as the direction of the acceleration. For objects in uniform circular motion, the net force and subsequent acceleration is directed inwards. It is often said that circular motion requires an inward (or "centripetal") force.

Without a centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion continues in motion in a straight line at constant speed.

Example: The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out.

(a) What force keeps the people from falling out the bottom of the cylinder?

Draw an FBD for a person with their back to the wall:

x

y

w

N

fs

It is the force of static friction.

(b) If μs = 0.40 and the cylinder has r = 2.5 m, what is the minimum angular speed of the cylinder so that the people don’t fall out?

Apply Newton’s 2nd Law:

( )( ) 0 2

1 2

=−=

===

∑∑

wfF

rmmaNF

sy

rx ω

( )

( )( ) rad/s 13.3m 5.240.0

m/s 8.9 2

2

===∴

==

=

rg

mgrmN

wf

s

ss

s

μω

ωμμ

From (2): From (1)

Example continued:

Example: A coin is placed on a record that is rotating at 33.3 rpm. If μs = 0.1, how far from the center of the record can the coin be placed without having it slip off?

Apply Newton’s 2nd Law:

x

y

w

N

fs

Draw an FBD for the coin:

( )( ) rmmgNf

rmf

sss

s2

2 :1 From

ωμμ

ω

===

=

From (2)

Solving for r:

( )( )( )

m 08.0rad/s 50.3

m/s 8.91.02

2

2 ===∴ωμ gr s

2ωμ gr s=

What is ω?rad/s 5.3

sec 60min 1

rev 1rad 2

minrev3.33 =⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=πω

( )( ) 0 2

1 2

=−=

===

∑∑

wfF

rmmaNF

sy

rx ω

Unbanked and Banked CurvesExample (text problem 5.20): A highway curve has a radius of 122 m. At what angle should the road be banked so that a car traveling at 26.8 m/s has no tendency to skid sideways on the road? (Hint: No tendency to skid means the frictional force is zero.)

θ

Take the car’s motion to be into the page.

FBD for the car:

x

y

N

w

θ

Apply Newton’s Second Law:( )

( ) 0cos 2

sin 12

=−=

===

∑∑

wNFrvmmaNF

y

rx

θ

θ

Rewrite (1) and (2): ( )

( ) mgNrvmN

=

=

θ

θ

cos 2

sin 12

Divide (1) by (2):

( )( )( )

°=

===

0.31

6007.0m 122m/s 8.9

m/s 8.26tan 2

22

θ

θgrv

rvmam

rMGmFF srs

esg

2

2 ====∑

Consider an object of mass m in a circular orbit about the Earth.

Earth

rThe only force on the satellite is the force of gravity:

rGMv

rvm

rMGm

e

ses

=

=2

2Solve for the speed of the satellite:

Circular Orbits

Example: How high above the surface of the Earth does a satellite need to be so that it has an orbit period of 24 hours?

rGMv e=From previous slide:

Also need,T

rv π2=

Combine these expressions and solve for r:3

12

24⎟⎠⎞

⎜⎝⎛= TGMr e

π

( )( )( )

m 10225.4

s 864004

kg 1098.5/kgNm1067.6

7

31

22

242211

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××=

−

πr

km 000,35=−=⇒+= ee RrhhRr

Example: What is the minimum speed for the car so that it maintains contact with the loop when it is in the pictured position?

FBD for the car at the top of the loop:

N w

y

x

Apply Newton’s 2nd Law:

rvmwN

rvmmawNF ry

2

2

=+

−=−=−−=∑

r

The apparent weight at the top of loop is:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=+

grvmN

rvmwN

2

2

N = 0 when

grv

grvmN

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−= 0

2

This is the minimum speed needed to make it around the loop.

Consider the car at the bottom of the loop; how does the apparent weight compare to the true weight?

N

w

y

x

FBD for the car at the bottom of the loop:

Apply Newton’s 2nd Law:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=−

==−=∑

grvmN

rvmwN

rvmmawNF cy

2

2

2