Examples Of Central Forces
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Examples Of Central Forces

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Central Forces

Central Forces

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Examples Of Central Forces Examples Of Central Forces Presentation Transcript

  • If the force on a body is always towards a fixed point, it is called a central force. Take the fixed point as the origin. Examples of central forces 1. uniform circular motion 2. force due to gravitation 3. simple harmonic motion 4. projectile motion 5. uniformly accelerated motion 6. others, like electrostatic , magnetostati c forces, etc.
  • — By studying central for ces you may master — 1. uniform circular motion 2. force due to gravitation 3. simple harmonic motion 4. projectile motion 5. uniformly accelerated motion at the same time ! —All
  • Since forces involve mass and acceleration , acceleration involves differentiation of velocity, velocity is differentiation of displacement, we need to know differentiation prior to it. it. Since displacement, velocity, acceleration and force are vector quantities, we need to know vectors prior to it. Then what we are required to know is it. vectors, vectors , differentiation and vector differentiation of course . course.
  • differentiation of vector functions of scalar variable- variable - time in Cartesian coordinates vector r of a moving mass point may be resolved (Position — into x and y components in Cartesian coordinates as r cos q and r sin q respectively. We write r = x + y = r cos q i + r sin q j .(1) — where i and j are unit vectors in x and y directions — respectively. On differentiation, we get, — or , v = vx + vy ... .(2) — where vx and vy as respectively and velocity is vector — differentiation of position vector.
  • DIFFERENTIATION OF VECTORS CARTESIAN COORDINATES (CONTINUED FROM PREVIOUS SLIDE) ¢ where vx and vy as respectively and velocity is vector differentiation of position vector. ¢ Eqn.(2) makes an important statement that the components of velocity in Cartesian coordinates are time derivatives of the components of position vectors. This result appears too obvious, but as we would see later, it may not hold in other system of coordinates .A second differentiation gives ¢ or , a = ax + ay . .(3)
  • DIFFERENTIATION OF VECTORS CARTESIAN COORDINATES (CONTINUED FROM PREVIOUS SLIDE) dv y dv x where ax and ay are respectively and dt dt or 2 a n d respectively as d 2x d 2y dt dt 2 acceleration is vector differentiation of velocity vector. Eqn.(3)similarly states that the components of acceleration in Cartesian coordinates are time derivatives of the components of velocity vectors. Again it may not hold in other system of coordinates.
  • DIFFERENTIATION OF VECTORS POLAR COORDINATES Y Q ds dr P r+dr dq q Y q p/2+q s r y T r q q X O x X R Fig 1:Resolution of radius vector into components
  • DIFFERENTIATION OF VECTORS POLAR COORDINATES Instead of di fferentiating displacement and vel ocity ¢ vectors, let us differentiate unit vectors r and Ù Ù (taken to each other) . Expressi ng them in Cartesian coordinates, or resol ving into components ¢ r =cos q i + sin q j and = - sin q i + cos q j .(5) Ù Ù ¢ Since magnitudes of both of them uni ty but directions are both vari ables . (see the figure in the above sl ide, no 7. ¢ For differentiation of the unit vectors refer to the figure in the next slide. Later on the formul a for differentiation of unit vectors shal l be fruitfully utilised for differentiating displacement and vel ocity vectors.
  • The unit vectors , , thei r increments Ù Ù Ù Ù dr dr r q ,are shown in the figure. Ù q Q S Ù q Ù Ù q+ d q Ù Q dq Ù q Ù Ù P r+ dr Ù Ù r+ dr q Ù r T q q P Ù Ù j q+ d q Ù r A A O P r=1 O S x Fig 2 : differentiation of unit vectors
  • DIFFERENTIATION OF UNIT VECTORS. as the unit vector makes an angle q with the x axis and the unit ¢ vector makes an angle p/2+q with the x axis and both the unit Ù Ù vectors have obviously magnitudes unity. Mind it that r and are unit vectors continuously changing in direction and are not constant vectors as such; whereas i and j are constant vectors. ¢ Differentiating the unit vectors with respect to time t, we have,(from Ù Ù (5) above) d r = - sin q dq i + cos q dq j and d = - cos q dq i - sin q dq j respectively dt dt dt dt dt dt Ù Ù or, d r = (- sin qi + cos qj) dq = Ù dq and respectively, dq dq Ù = (- cos qi - sin qj ) = - r ¢ d dt dt dt dt dt dt Ù dr Ù Ù or and = - r wrespectively .. .(6) Ù =w d ¢ dt dt dq w= where , the magnitude of angular velocity of the moving ¢ dt particle around the point O, or the time rate of turning of q . Ù dr It is important to see here that dt is parallel to , i.e., Ù ¢ Ù perpendicular to r , i.e., in a direction tangent to the unit circle. ¢ Ù Also d q is parallel to - r , i.e., along the radius and towards the Ù Ù dt d2 r Ù center, and thus it is perpendicular to . Thus dt2 is parallel ¢ Ù to d q , i.e., parallel to - r . Ù ¢ dt Ù Ù Thus the derivative of is in the direction of - r or centripetal. ¢
  • DIFFERENTIATION OF VELOCITY AND ACCELERATION VECTORS
  • WHAT IF THE FORCE IS ALWAYS TOWARDS A FIXED POINT, I.E., CENTRAL FORCE
  • Different cases of central force æ .. öÙ æ. öÙ . m ç r - r w2 ÷ r + m ç 2 r w + r w ÷ Ù F = ma, then Frr + F = è ø è ø .. 1. For uniform circular motion, r =a, is a constant and r = 0 since r is a constant. So F r = - a w 2 F =0 .. 2. For simple harmonic motion, F =0, =0, r = - kr 3. For projectile motion, simpler will be Cartesian coordinates, ax =0, and ay =-g, and uniform acceleration is a particular case of projectile motion where the horizontal velocity is 0 always.