Download presentation

Presentation is loading. Please wait.

Published byLuis Quinlan Modified over 2 years ago

1

2
Circular Motion When an object travels about a given point at a set distance it is said to be in circular motion

3
Cause of Circular Motion 1 st Law…an object in motion stays in motion in a straight line at a constant speed unless acted on by an outside force. 2 nd Law…an outside (net) force causes an object to accelerate in the direction of the applied force. THEREFORE, circular motion is caused by a force acting on an object pulling it out of its inertial path in the direction of the force.

4
Circular Motion Analysis r v1v1 r v2v2 q

5
v1v1 v2v2 r r v1v1 v2v2 v = v 2 - v 1 or v = v 2 + (-v 1 ) (-v 1 ) = the opposite of v 1 v1v1 (-v 1 )

6
v1v1 v2v2 r r v2v2 (-v 1 ) v = v 2 - v 1 or v = v 2 + (-v 1 ) (-v 1 ) = the opposite of v 1 v1v1 (-v 1 ) v1v1 v2v2 v Note how v is directed toward the center of the circle q

7
v1v1 v2v2 r r v1v1 v2v2 v2v2 (-v 1 ) v l Because the two triangles are similar, the angles are equal and the ratio of the sides are proportional

8
v1v1 v2v2 r r v1v1 v2v2 v2v2 (-v 1 ) v l Therefore, v/v ~ l/rand v = v l/r now, if a = v/tand v = v l/r then, a = v l/rtsince v = l/t THEN, a = v 2 /r

9
Centripetal Acceleration a c = v 2 /r now, v = d/t and, d = c = 2 r then, v = 2 r/t and, a c = (2 r/t) 2 /r or, a c = 4 2 r 2 /t 2 /r or, a c = 4 2 r/T 2

10
The 2 nd Law and Centripetal Acceleration FcFc acac vtvt F = ma a c = v 2 /r = 4 2 r/T 2 therefore, F c = mv 2 /r or, F c = m4 2 r/T 2

11
Motion in a Vertical Circle A B FwFw TATA FwFw TBTB

12
Vertical circle FwFw TATA FwFw TBTB A B Top of Circle at v min T A = 0 and F w = F c therefore, T A + mg = mv 2 /r because T A = 0, mg = mv 2 /r and v 2 = rg

13
Vertical Circle FwFw TATA FwFw TBTB A B Bottom of Circle v max at bottom therefore, T B + mg = mv 2 /r F c = T B + F w or or, T B = mv 2 /r - mg

14
Cornering on the Horizontal When an object is caused to travel in a circular path because of the force of friction, then,... F c = F F car FwFw FNFN F

15
Cornering on the Horizontal F c = F F car FwFw FNFN F Therefore, mv 2 /r = F N on horiz., F N = F w = mg, mv 2 /r = mg … or, = v 2 /rg

16
``````` Cornering on a Banked Curve car FwFw FNFN FcFc FcFc

17
Cornering on a Banked Curve ` car FwFw FNFN FcFc FNFN FcFc Note how F N is the Resultant FwFw

18
FwFw FNFN FcFc If we want to know the angle the curve has to be at to allow the car to circle without friction, then we have to analyze the forces acting on the car. Sin = F c /F N F c = Sin F N F c = mv 2 /r Sin F N = mv 2/ r therefore, mv 2 /r= Sin F N

19
FwFw FNFN FcFc Sin = F c /F N F c = Sin F N F c = mv 2 /r Sin F N = mv 2/ r therefore, mv 2 /r = Sin F N Cos = F w /F N F N = F w /Cos F w = mg F N = mg/Cos mv 2 /r = Sin F N F N = mg/Cos or, mv 2 /r Sin = mg/Cos tan = v 2 /rg

20
FCFC FWFW FNFN Note! F N is resultant FCFC FWFW FNFN Cos = F W /F N and Sin = F C /F N F N = F W /Cos mg/Cos and F N = F C /Sin mv 2 /r Sin mg/Cos = mv 2 /r Sin Sin /Cos = mv 2 /rmg Tan = v 2 /rg F N supplies F C for circular motion, no F F needed OR

21
Universal Gravitation E M Ah, the same force that pulls the apple to the ground pulls moon out of its inertial path into circular motion around the earth! Therefore, the forces must be proportional to each other!

22
E M Now, if the earth pulls the apple at a rate of 9.8 m/s 2, then, the same earth must pull to moon at a proportional rate to that. 60 re If the moon is 60 x further from the apple, and all forms of energy obey the Inverse Square Law, then, the acceleration of the moon should be 1/60 2 of that of the apple, or 9.8 m/s 2 x 1/60 2 = m/s 2 And, in one second it should fall d = 1/2 (0.0072m/s 2 )(1sec) 2 or, m

23
Universal Gravitation FEFE Force of Earth on moon FMFM Force of Moon on Earth F E = F M 3 rd Law

24
Universal Gravitation FEFE Force of Earth on moon FMFM Force of Moon on Earth F E = F M 3 rd Law Because of the 3 rd Law and the Inverse Square Law : F = Gm 1 m 2 /r 2

25
Universal Gravitation F = Gm 1 m 2 /r 2 If F is the weight of an object, F w, then, F w = m 2 g and, m 2 g = Gm 1 m 2 /r 2 or, g = Gm 1 /r 2 or, m 1 = gr 2 /G

26
Universal Gravitation If gravity is the force that causes an object to travel in circular motion, then, F = F c or, Gm 1 m 2 /r 2 = m 2 v 2 /r or, m 1 = v 2 r/Gor,r = Gm 1 /v 2 or, v 2 = Gm 1 /r or, m 1 = v 2 r/G

27
Universal Gravitation If gravity is the force that causes an object to travel in circular motion, then, F = F c or, Gm 1 m 2 /r 2 = m 2 v 2 /r or, Gm 1 m 2 /r 2 = m r/T 2 transpose extremes T 2 /r 2 = m r/Gm 1 m 2 divide by r and cancel m 2 T 2 /r 3 = 4 2 /Gm 1

28
Universal Gravitation T 2 /r 3 = 4 2 /Gm 1 Note that for any object circling a superior object that 4 2 /Gm 1 remains constant!!!! Therefore, T 2 /r 3 is also constant for all objects circling that superior object

29
Universal Gravitation T 2 /r 3 = k for all circling objects Therefore,for two objects circling the same superior object... T 1 2 /r 1 3 = T 2 2 /r 2 3 or (T 1 /T 2 ) 2 /(r 1 /r 2 ) 3

30
Keplers Laws 1 st Law…all planets circle the Sun in ellipital paths with the Sun at one focus Sun planet F2F2 F2F2

31
Keplers Laws 1 st Law…all planets circle the Sun in ellipital paths with the Sun at one focus 2 nd Law…Each planet moves around the sun in equal area sweep in equal periods of time

32
Keplers Laws 2 nd Law…Each planet moves around the sun in equal area sweep in equal periods of time b a Area 12a = Area 43b

33
Keplers Laws 1 st Law…all planets circle the Sun in ellipital paths with the Sun at one focus 2 nd Law…Each planet moves around the sun in equal area sweep in equal periods of time 3 rd Law…the ratio of the squares of the periods to the cube of their orbital radii is a constant

34
Keplers Laws 3 rd Law…the ratio of the squares of the periods to the cube of their orbital radii is a constant T 2 /r 3 = k for all circling objects Therefore,for two objects circling the same superior object... T 1 2 /r 1 3 = T 2 2 /r 2 3 or(T 1 /T 2 ) 2 /(r 1 /r 2 ) 3

35
Sample Problems What is the gravitational attraction between the Sun and Mars? F = ? m s = 1.99 x kg m m = 6.42 x kg r m = 2.28 x m F = Gm s m m /r m 2 F = 6.67 x N m 2 /kg 2 (1.99 x kg)(6.42 x kg) (2.28 x m) 2 F = 1.64 x N

36
Sample Problems What velocity does Mars circle the Sun at? v = ? m s = 1.99 x kg m m = 6.42 x kg r m = 2.28 x m F = F c Gm s m m /r m 2 = m m v 2 /r m v 2 = Gm s /r v 2 = 6.67 x Nm 2 /kg 2 (1.99 x kg)/2.28 x m v = 2.4 x 10 4 m/s

37
Sample Problems What is the period of Mars as it circles the Sun? T = ? m s = 1.99 x kg m m = 6.42 x kg r m = 2.28 x m F = F c Gm s m m /r m 2 = m m 4 2 r/T 2 T 2 = 4 2 r 3 /Gm s T 2 = 4 2 (2.28 x m) 3 /6.67 x )1.99 x kg T = 5.9 x 10 7 s or,T = 685 days

38
Sample Problems What is the period of Mars? This time use Keplers 3 rd Law to find it! T m = ? T e = da r e = 1.5 x m r m = 2.28 x m F = F c Gm s m m /r m 2 = m m 4 2 r/T 2 T 2 /r 3 = 4 2 /Gm s T m 2 /r m 3 = T e 2 /r e 3 T m 2 /(2.28 x m) 3 = ( da) 2 /(1.5 x m) 3 T m = 684 days

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google