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- 1. 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 , magnetostatic forces, etc.
- 2. By studying central forces 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
- 3. Since forces involve mass and acceleration, acceleration involves differentiation of velocity, velocity is differentiation of displacement, we need to know differentiation prior to 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 vectors, differentiation and vector differentiation of course.
- 4. differentiation of vector functions of scalar variable- time in Cartesian coordinates vector r of a moving mass point may be (Position resolved into x and y components in Cartesian coordinates as r cos and r sin respectively. We write r = x + y = r cos i + r sin 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.
- 5. 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)
- 6. DIFFERENTIATION OF VECTORS CARTESIAN COORDINATES (CONTINUED FROM PREVIOUS SLIDE) dv y dv x where ax and ay are dt a n d dt respectively or 2 2 respectively as y dx d and 2 2 acceleration t vector differentiation of dt d is 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.
- 7. DIFFERENTIATION OF VECTORS POLAR COORDINATES Y Q s r P r+ r Y s /2+ r y T r X O x X R Fig 1:Resolution of radius vector into components
- 8. DIFFERENTIATION OF VECTORS POLAR COORDINATES Instead of differentiating displacement and velocity vectors, let us differentiate unit vectors and θ r (taken ┴ to each other) . Expressing them in Cartesian coordinates, or resolving into components i + sin j and θ= - sin i + cos j ….(5) r =cos Since magnitudes of both of them unity but directions are both variables . (see the figure in the above slide, no 7. For differentiation of the unit vectors refer to the figure in the next slide. Later on the formula for differentiation of unit vectors shall be fruitfully utilised for differentiating displacement and velocity vectors.
- 9. The unit vectors , , their increments r r r ,are shown in the figure. Q S Q P r r r r r T P r A’ A O P r=1 O S x Fig 2 : differentiation of unit vectors
- 10. DIFFERENTIATION OF UNIT VECTORS. as the unit vector makes an angle with the x – axis and the unit vector makes an angle /2+ with the x – axis and both the unit vectors have obviously magnitudes unity. Mind it that and are unit vectors θ r 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 (5) above) d r sin d i cos d j and d θ cos d i sin d j respectively dt dt dt dt dt dt or, d r dθ and respectively, d d d d i cos j θ cos i j r sin sin dt dt dt dt dt dt dr dθ and respectively…………………..……….(6) θ or r dt dt d where , the magnitude of angular velocity of the moving particle dt around the point O, or the time rate of turning of . dr It is important to see here that is parallel to , i.e., θ dt perpendicular to , ri.e., in a direction tangent to the unit circle. Also d is parallel to , i.e., along the radius and towards the r dt dr 2 θ Thus center, and thus it is perpendicular to . is parallel 2 dt to d , i.e., parallel to r. dt Thus the derivative of is in the direction of orr centripetal. θ
- 11. DIFFERENTIATION OF VELOCITY AND ACCELERATION VECTORS
- 12. WHAT IF THE FORCE IS ALWAYS TOWARDS A FIXED POINT, I.E., CENTRAL FORCE
- 13. Different cases of central force .. . . r θ 2 mr r m 2r r F = ma, then Frrˆ + Fθ = θ .. 1. For uniform circular motion, r =a, ω is a constant and r 0 since r is a constant. So F rˆ = - a 2 Fθ=0 .. 2. For simple harmonic motion, Fθ=0, ω =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.

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