Advanced Higher Physics Unit 1 Angular motion. Many motions follow a curved path. v w θ θ angular...

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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
Example2007 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.
Example2007 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=vu 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 stringvertical 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
Example2006 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