NON-uniform Circular Motion

1. * The NET acceleration is no longer pointing towards the centre of the circle. Radial acceleration Tangential acceleration centre NET acceleration NON-uniform Circular Motion * There are TWO components of acceleration: Radial / centripetal : due to the change in direction of velocity Tangential : due to the change in magnitude of velocity Speed is changing

2. Roller Coaster Vertical circle with a string and bob string v w bob Examples of non-uniform circular motions

3. Change in direction T q Change in speed mg Free body diagram Vertical circle with a string and bob Radial direction : T - mg cos q = mac = mv2 / r string q Tangential direction : mg sin q = mat bob mg sin q mg cos q

4. Can an object (mass m) go round a vertical circle of radius l if the initial speed at the bottom is u? C Can go round the circle : (1) Have enough energy to reach point C. (2) Have sufficient high centripetal force to maintain the circular motion at C. D B Consider Conservation of energy ; l m u A  0

5. v T mg Consider force at point C ; By Conservation of energy, Can an object (mass m) go round a vertical circle of radius l if the initial speed at the bottom is u? C Can go round the circle : (1) Have enough energy to reach point C. (2) Have sufficient high centripetal force to maintain the circular motion at C. D B l m u  0 A

6. The object can go round the circle if the initial speed is greater than What happens if u < ? Can an object (mass m) go round a vertical circle of radius l if the initial speed at the bottom is u? C Can go round the circle : (1) Have enough energy to reach point C. (2) Have sufficient high centripetal force to maintain the circular motion at C. D B l m u A

7. (1) < u < Can reach C (as u > ) No more circular motion can be processed (as T = 0 but mg is greater than mv2/l) Projectile motion due to gravity What happens if u < ? C D B l m u A

8. (2) < u < Between B and C(as u < ) Projectile motion due to gravity (3) u < Cannot reach B For reaching B, 1/2 mu2 = 1/2mvB2 + mgl u2 2gl u  Swing about A between B and D What happens if u < ? C D B l m u A

9. R man R = 0 for weightless Consider the whole system (spaceship and man), mg v R Consider the man only, R mg Mg r More about Circular Motion * A astronaut feels weightless in a spaceship which is moving with uniform circular motion about the Planet, say the Earth. Mg + mg = (M+m) v2 / r v2 = g r mg -R = mv2 / r mg -R = m(g r) / r mg -R = mg R = 0

10. R R = mg’ mg’ man Rotating axis w R r w More about Circular Motion * Artificial gravity made for Space stations No weight as it is far away from all planets There is only normal contact reaction force due to contact N.

11. More about Circular Motion * Working principle of a centrifuge

12. The pressure gradient increases with the distance from the rotating axis FC = P A = (P2 - P1 )A = mrw2 * Working principle of a centrifuge (1) Assume it is horizontally aligned with liquid of density r inside. P1 = P P2 = P+P (P2 - P1)A Pressure gradient as centripetal force

13. Net force = (P2 - P1 )A * Working principle of a centrifuge (2) Consider an element of the liquid ofdensity r inside. All liquid rotates with uniform speed Net force due to pressure gradient = r r A w2 r = [m] r w2 = [rV] r w2 = r(Ar) r w2

14. Move towards the axis Net force Fnet= (P2 - P1 )A = r r A w2 r Move away from the axis * Working principle of a centrifuge (2) Consider an element of other substance ofdensity r’ inside. r’ r’< r for less dense object Required centripetal force Fc = [m’] r w2 = [r’V] r w2 r’> r for denser object = r’(Ar) r w2 = r’r A w2 r

15. FC= r’r A w2  r Fnet = r r A w2  r ~ 700 / 1 More about Circular Motion * Why centrifuge ? Excess force for separation  Fg = weight - upthrust = (r’ A r g ) - (r A  r g) = (r’ - r) A g  r Assume r’ > r Excess force for separation  Fc = (r’ - r)r A w2  r Typical : r = 10 cm, w = 2500 rev min-1