When a mass moves because of a force, we can define several types of problem.

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Sydney Fisher
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1 Mechanics Lectue 4 3D Foces, gadient opeato, momentum 3D Foces When a mass moves because of a foce, we can define seveal types of poblem. ) When we know the foce F as a function of time t, F=F(t). ) When we know the foce F as a function of position vecto. This is a much moe common poblem and we shall discuss it in detail. We have aleady solved poblems of this type such as motion in a Hooke s law foce (e.g. F=-x). 3) When we know the foce F as a function of position vecto and velocity. Will solve a D poblem of this type this week involving the motion of a dopped ball with ai esistance (F=mg-b ) (Pactice session 4, poblem 5). Hee we shall mainly conside case ) above because it is vey impotant fo solving a lot of poblems such as the motion of planets o oscillation poblems. Fist let us conside an example of case ) whee F=F(t). Such poblems ae vey easily solved by diect integation. Example The foce on a mass m in the x diection inceases with time accoding to the equation Fx=At, whee A is a constant. What is the position of the mass as a function of time? This is easy to solve by integation of Newton s nd law: mx Fx At. This is a mass with an acceleation that inceases linealy with time. We expect the mass to get faste and faste vey quickly. By integation,, whee B is a constant, and x At / m B, 6 3 x At / m Bt C whee C is a constant. The constants B and C can be found fom initial conditions.

2 In this lectue we will be mainly consideing the case ) above whee F depends on only. These poblems ae a little bit moe difficult to solve, but we will soon see how. Poblems concening case 3) in thee dimensions will not be dealt with in this couse. 3D motion in a foce field, Wok enegy theoem Conside a point mass m moving in thee dimensions (3D) unde the influence of a foce field F, that is F=F(), with no fiction (that is, no foces depending on velocity). Thee ae no extenal foces apat fom F. The kinetic enegy is given by ( v v x y z ) T mv m m v v v By diffeentiating, dt dt m d v v F v. dt This is the powe being deliveed to mass m by the foce F. As the mass m moves, F changes. It is useful to calculate the change in kinetic enegy when the mass moves fom one position to anothe position. dt dt d F dt F. d dt So, T T F d But we also know that the wok done by F is W F. d. Fo shothand, we wite T T T, so T F. d W ( ), ()

3 whee W ( ) is the wok done by F when the paticle moves fom position to position. This is the wok-enegy theoem in 3D (see Lectue 3 p. 3 fo the D vesion of this poblem). We can wite F. d Ft d, whee Ft is the tangential component of F (the component of F along the diection of motion.) We can also wite Eq. () using T mv m v change in v fom ( v ) v v. d m F., so that we can find the A consequence of dw F. d is that no wok is done if F is at ight angles to d, as in the following example of hoizontal cicula motion of a mass on a light sting: Foce with which the sting pulls the mass F m m (see Lectue ). ˆ Theefoe W( ) F. d 0 3D Consevative foces Definition of a consevative foce fo a single paticle of mass m: (i) F=F() (not depending on v o diectly o on t) (ii) W ( ) is the same fo all outes 3

4 In Lectue, p. 4 we defined, fo a slow change, potential enegy change =wok done by an extenal foce Fe potential enegy change = ( wok done by the system foce F) because F=-Fe. So, in 3D, fo a slow change in moving the paticle fom a position 0 fo which V=0 to position is given by V ( ) W ( ) F( ). d whee W is the wok done by system foce F 0 0 (0, the position in which V=0, can sometimes be at =.) In the diagam below we label the points 0, and meaning the position vectos ae at 0,, and. V F ( ). d and V F ( ). d, 0 0 so V V F( ). d F( ). d. 0 0 But we know that fo a consevative foce, W ( ) is the same fo all outes. Theefoe oute0, 0 oute So V V F( ). d fo a slow change If the change is not slow, i.e., fo motion of a paticle in field F with no extenal foce, V V F( ). d ( T T ) fom the wok-enegy theoem (Eq. on page ). V T V T E mechanical enegy. So E=T+V at any position. 4

5 The gadient opeato Conside a small displacement d of a mass m in a consevative foce field F(). We can calculate the enegy change in two ways: Method : dw F. d Fx dx Fydy Fz dz Method : Use V position whee V=0). Theefoe dv W (whee W is the wok done by F when m moves fom the dw dw dv V ( x dx, y dy, z dz) V ( x, y, z) V V V dx dy dz x y z If we compae method and method, we should get the same answe, so V x, V F Fy, F V z x y z, o V x V F V gadv y fo a consevative foce field. V z This implies that dw V. d. GMm Example Conside the potential enegy V ( x, y, z), whee G, M and m ae constant and. Find F. (This is in fact a gavitational potential enegy. See diagam.) F V GMm. 5

6 We can find as follows: (/ ) x (/ ) y. (/ ) z But (/ ) ( x y z ) x x x / x x / 3 ( x y z ). x y 3 3. z GMm GMmˆ F, whee 3 ˆ is the unit vecto in the diection of. So we have found the invese squae law of gavity. (In fact G.) Nm kg Undestanding the meaning of V V x dx V V V V dv dx dy dz dy V. d x y z y. dz V z This equation shows that dv=0 when d is pependicula to V. Conside fist the example of two-dimensions (D) with x and y only. On a plot of V(x, y) x, y we can daw contous of equal V. 6

7 Moving in the diection pependicula to the contous allows V to change the most. So dv is lagest and positive when d points up the slope. But since dv V. d, then dv is the lagest when V and d ae paallel. Theefoe V must point up the slope, pependicula to the contous of constant V. The diection and magnitude of field. V depend on position so we call V a vecto The diagams above ae fo the example of D. You can see that pependicula to the contous, and towads highe V. In 3D, the diection of inceasing V: V V (gadv) points also points towads 7

Example Sketch the vecto field V GMmˆ example). We know that V. GMm when V ( x, y, z) (as in the pevious The field of F is opposite to the field of V because F V. Path independence and loops F( ).

8 Example Sketch the vecto field V GMmˆ example). We know that V. GMm when V ( x, y, z) (as in the pevious The field of F is opposite to the field of V because F V. Path independence and loops F( ). d F( ). d F( ). d ( A) ( B) ( B) F( ). d F( ). d 0 ( A) ( B) F ( ). d 0. No pepetual motion machines can be built with this type of foce field because we cannot get wok out by moving an object in a loop. 8

9 Rotational motion Angula momentum of a paticle L p in kg m s - Visualizing angula momentum ) L psin pt, whee pt is the tangential component of vecto p. ) L sin p p p whee p is the distance shown below. 3) x px 0 L y p 0 ( ) ˆ y xpy ypx k 0 0 xpy yp x 9

10 Example Cicula motion L p ˆ, p mv me ˆ, whee is the instantaneous tangential diection. So L m ˆ eˆ m kˆ m k ˆ. ê The angula momentum is in the +z diection if the sense of the otation is anticlockwise about the z axis. This is as in a ight hand cokscew. In this case T mv m. If the mass is held by a light sting, the tension in the sting is mv F m m. Example Motion in a staight line ) On the x axis L ˆi mvˆi 0 ) Paallel to the x axis 0

11 x mv 0 L a 0 0 mav ˆ k. 0 0 mav The angula momentum is in the z diection (fo positive v and a) because the sense of otation about the oigin is clockwise. Toque The toque is given the symbol G, and is defined by G F Ft pf whee the subscipts t and p have the same meaning as befoe on p. 9. Fom Newton s nd law, we know that, so F p G p d But L ( p) p p m p 0 p p dt Theefoe G L This is Newton s nd law ewitten to apply to angula motion The units of G ae Nm, in fact the same as enegy because Nm= J. The physical meaning of G is the equivalent of a foce fo otational motion. The stonge the toque the stonge the powe of otation. Fom G F we can see that the diection of G is pependicula to and to F. The diection can be found fom the sense of otation in the same way as we found the diection of L fom and p. dl If G=0 then 0 dt constant. L=constant, and the angula momentum of the paticle is (Please do not confuse toque G, a vecto, with the gavitational constant G.)

Quiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)

Quiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible) Name: Class: Date: ID: A Quiz 6--Wok, Gavitation, Cicula Motion, Toque. (60 pts available, 50 points possible) Multiple Choice, 2 point each Identify the choice that best completes the statement o answes