Advanced Higher Physics Unit 1

Angular motion

Angular motion

Many motions follow a curved path.

vwθ

θ angular displacement ,measured in radians (rad)

w angular velocity, measured in radians per second (radsˉ¹)

α angular acceleration, measured in radians per second per second (radsˉ²)

v rotational (tangential) velocity

a rotational (tangential) acceleration

Angular displacement

rador2

90

rador 180

rador 2

3270

rador 2360

rad2

rad

rad2

3

rad2

Angular velocity

Angular velocity w is defined as the change in angular displacement dθover time dt.

dt

dw

wdθ

Example-2007 Q2(a)(ii)(A)

260

revsofnb

w

260

45w

Axis of rotation

A turntable accelerates uniformlyfrom rest until it rotates at 45 revolutions per minute. The time taken for the acceleration is 1.5 s.

Show that the angular velocity after 1.5 s is 4.7 rad.sˉ¹.

Solution

17.4 sradw

Angular acceleration

Angular acceleration α is defined as the change in angular velocity dwover time dt.

dt

dw

As with linear motion:

2

2

dt

d

dt

dw

(This formula can be found in the data booklet)

Equations of angular motion

The equation of angular motion are similar to those of linear motion.

tww 0

Angular motion Linear motion

20 2

1ttw

220

2 ww

tww

2

0

atuv 2

2

1atuts

atuv 222

tvu

s

2

You do not need to derive these !

w◦ initial angular velocity, measured in radsˉ¹.

Example-2007 Q2(a)(ii)(B)

17.4 radsw10 radsw

?

Axis of rotation

A turntable accelerates uniformlyfrom rest until it rotates at 45 revolutions per minute. The time taken for the acceleration is 1.5 s.

Calculate the angular acceleration of the turntable.

Solution

st 5.1rad280245

tww 0

5.107.4

5.1

7.4

11.3 rads

Uniform Circular Motion

2

Tt

Tw

2

For this motion, both w and v are constant.T is the period of the motion, that is the time for one complete rotation.

θ

v

s

vLinear motion

wT

2

Angular motion

rs 2Tt

T

rv

2

v

rT

2

v

r

w

22

rwv (In data booklet)

You need to be able to derive this!

Also:

rs

ra

(These formulas can be found in the data booklet)

Radial acceleration

v

v 2

1

sin

θ

v

u

A

B

θ

A particle moves from A to B in time Δt with a constant speed.

Its velocity is changing (size stays the same, but direction changes).

The change in velocity Δv=v-u is:

-u v

Δv

θ

sin2vv

θ

v

u

A

B

θ s

v

rt

2

v

st

r

rs t

va

vrv

a

2sin2

r

va

sin2

but

so

As θ → 0, then a → instantaneous acceleration

r

va

sinlim

2

0

sinlim

0

2

r

va

r

va

2

or2rwa rwv since

(can be found in the data booklet)

You need to be able to derive these!

This acceleration is towards the centre of the circle.

Any circular motion must have a radially inwards force responsiblefor the motion (F=ma).

This force is called the CENTRIPETAL FORCE.

22

mrwr

mvF

Centripetal Force

(In data booklet)

Examples: ORBIT

F

The centripetal force is supplied by thegravitational force.

Car on a track

The centripetal force is supplied by Friction.

Car at the top of a bump eg: bridge

The centripetal force is supplied by weight.

Mass on a string-vertical circle

mgTF

mgTF

The centripetal force is supplied bya combination of tension and weight.

TF

Conical pendulum

WT cos

FT sin

θ

T

F

The centripetal force F is supplied by component of tension T.

W

mg

Ftan

Example-2006 Q1(a)(ii)(iv)(v)

Y

XA child’s toy consist of a model aircraft attached to a light cord. The aircraft is swung in a vertical circle at constant speed as shown.X is the highest point and Y is the lowest point in the circle.

Time for 20 revolutions: 10.00sRadius of circle: 0.500mMass of aircraft: 0.200kg

1. Calculate the centripetal force acting on the aircraft.

2. Draw labelled diagrams to show the forces acting on the aircraft at X and at Y.

3. Calculate the minimum tension in the cord.

Solutions

sT 5.020

10

mr 5.0

kgm 2.0

1.

radT

w 56.1245.0

2

2mrwF 2)4(5.02.0 F

NF 8.15

Solutions

2. At X

weight tension

At Y

weight

tension

Solutions

TWF WFT

mgFT

3. The minimum tension happen at X

At X

8.92.08.15 TNT 8.13