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P v A particle P travels at constant speed v around a circle of radius r r
2 P v r Speed is constant. Why does P have an acceleration? P Velocity is changing … in direction
3 360° 1 radian r q
1 r p radians = 180 °
4 q r s s
q =
r 5 P3 Find, (i) the number of radians in 17°, (ii) the number of degrees in 1 radian. (i) 180 ° = p radians 1 ° =
p
radians 180 p 17 ° =
x 17 180 = 0.297 radians (ii) p radians = 180 °
1 radian = 180 p
degrees 6 v P v = linear speed (or tangential speed)
7 time = t v P s distance speed = time s )
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w = rate of change of angle w.r.t. time v P w = o q x q
t Units of w = ?? Radians per second (rad s ­1 )
9 P5 If the angle traced out by P in 4 seconds is 10 radians, find its angular velocity. v w = P o q x q
t 10 ­ 1 ω =
= 2.5 rad s 4 10 P6 A small object moving in a circle at a steady speed does 3000 revolutions of the circle every minute. Find its angular velocity. 1 revolution = 2 p radians ­1 Þ 3000 rpm = 3000 x 2 p rad min 2p
3000 x 2 p
Þw =
= 314 rad s ­ 1 60 11 P7 A particle moving in a circle has a constant angular velocity of 2.2 rad s ­1 . What angle does it trace out in 6 s? θ ω = t θ 2.2 =
6 q = 13.2 rad
12 Relationship between Linear Speed and Angular Velocity
Points a, b, c … same w … different v v = rw 13 Proof of v = r w In t seconds … • • object traces out angle q travels distance s s
q =
r q s q s 1 =
= x t
rt t
t r 1 w = v x r v = w r 14 P 8 A wheel of radius 50 cm rotates with a constant angular velocity. If a point on the rim of the wheel has a speed of 10 m s ­1 , what is the angular velocity of the wheel? v ω=
r 10
=
= 20 rad s ­1 0.5 15 P 9 A stone moving at a steady speed in a circle of radius 2 m performs 10 complete revolutions of the circle per second. Find: (i) its angular velocity, (ii) its tangential or linear speed. (i) (ii) 1 revolution = 2p radians 10 revs. per second = 10 x 2p radians per second i.e. w = 20p rad s ­1 Linear speed = v = w r = 20p x 2 = 40p = 126 m s ­1
16 P 10 The radius of the Earth at the equator is 6.4 x 10 6 m. Assuming that the Earth rotates once abut its polar axis every 24 hours, find: (i) the angular velocity of the Earth, (ii) the linear speed of a point on the equator. Angle traced out (i) ω =
time taken =
2 π (24)(60)(6 0) = 7.27 x 10 ­ 5 rad s ­1
(ii) v = rω = (6.4 x 10 6 )(7.27x10 -5 ) = 465 m s ­1 17 Why does P has an acceleration? P
F = ma Þ Resultant force on P 18 P
Always perpendicular to velocity … why? Þ Towards the centre 19 mv 2
F = r = m ω 2 r P What does the magnitude of F depend on? m = mass of P v = linear speed r = radius of circle How does F depend on these ?? 20 P
• The force on • a particle moving in a circle • magnitude of F = mv 2 / r • direction of F is towards the centre 21 P
• The acceleration of • a particle moving in a circle • magnitude of a = v 2 / r • direction of a is towards the centre 22 P 11 A 2 kg stone moves in a circle of radius 3 m with a constant speed of 4 m s ­1 . Find: (i) its acceleration, (ii) the resultant force acting on it. v 2
(i) Centripeta l Accelerati on =
r mv 2
(ii) Resultant force to centre =
r OR
4 2 =
= 5.33 m s -1 3 (2)(4) 2
=
3 F = ma = 2 x 5.33 = 10.7 N = 10.66 N 23 P 12 A particle of mass 10 kg travelling at 8 m s ­1 is moving in a circular path with a constant angular velocity of 2 rad s ­1 . (i) What is the radius of the circle? (ii) What is the acceleration of the particle in magnitude and direction? 8
v =
= 4 m (i) v = r ω Þ r =
2 ω
(ii) a = r ω 2
= (4)(2)2 = 16 m s ­2 towards the centre 24 P 13 An electron in a magnetic field moves in a circular path of radius 1 cm. If its speed is 2 x 10 6 m s ­1 find the centripetal force on the electron. (Mass of electron = 9.1 x 10 ­31 kg) mv 2
F =
r (9.1 x 10 ­31 )(2 x 10 6 )
=
1 x10 ­ 2 ­16
= 3.64 x 10 N 25 v v S E v Circular Path Þ Centripetal Force d
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26 v v S E v v Centripetal Force mv 2
=
r G M m =
r 2 G M G M Þ v =
Þ v =
r r 2
What does v depend on?
27 P 14 A military spy satellite orbits the Earth at a height of 40,000 km above the Earth’s surface. At what speed is it travelling? (Given: G, radius and mass of Earth) G M v=
r (6.7 x 10 ­11 )(6.0x10 24 ) ­1 v = =
2943 m s 6.4 x 10 6 + 40,000x10 3 6.4 x 10 6 m 40,000 km = 40,000 x 10 3 m
28 T Period or Periodic Time v R Time to go once around distance time = speed 2 pR T =
v 29 P 15 What is the period of the satellite in P 14. (Its height above the Earth’s surface is 40,000 km and its speed is 2943.4 m s ­1 . ) 2 pR T =
v 2π (6.4 x 10 6 + 40,000 x 10 3 ) =
2943.4 T = 99049 s = 27.5 hours 6.4 x 10 6 m 40,000 km = 40,000 x 10 3 m
30 v R M m Circular Motion
Þ 3
Centripetal Force R T = 2π distance G Mm distance GM mv GM speed
=
Þ time Þ v = =
F ==
2
2
R time 2 pR 4π 2 R 2 2
T =
T =
v v 2 R 2
4π 2 R 2 T =
æ GM ö
ç
÷
è R ø
2
speed R 4π 2 R 3 T =
GM 2
31 R 3
T = 2π GM s
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32 P 16 The radius of the orbit of the planet Neptune about the sun is 30 times longer than the Earth’s radius of orbit around the Sun. Calculate the time it takes Neptune to complete one orbit of the Sun. æ 4π 2 R N 3 ö
çç
÷÷
2 3 2 3 2
T R 4π R GM T N
N
N 2
è
ø
=
T = Þ
=
2 3 2
2 3 T R E E GM T E æ 4π R E ö
çç
÷÷
è GM ø
2 N
2 E æ R N ö
T ÷÷
= çç
T è R E ø
3 = 303 T N2 = 30 3 1 T N = 164.3 year s 33 • T = 24 hours • Orbits in Equatorial Plane • Above same point on equator always • Same direction of rotation
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Satellite Earth
35 P 17 At what height above the Earth will a satellite in a parking orbit be found? 2 3 4π R 2
T =
GM 2 3 4π R 2
(24 x 60 x 60) =
(6.7 x 10 ­11 )(6.0 x 10 24 ) R = 4.24 x 10 7 m = (R E + h) h = (4.24 x 10 7 ­ R E ) = 3.6 x 10 7 m = 36,000 km
36 P 17 At what height above the Earth will a satellite in a parking orbit be found? Orbiting speed of satellite = ?? v =
GM R (6.7 x 10 ­11 )(6.0 x 10 24 )
v = 4.24 x 10 7 v = 3079 m s ­1 = 3.1 km s ­1
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