EP 222: Classical Mechanics - Lecture 34 Dipan K. Ghosh Indian Institute of Technology Bombay dipan.ghosh@gmail.com Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 34 October 29, 2014
Real & Pseudo Forces v r − � a f − � A ) − 2 � ˙ r − � Ω × ( � � a r = ( � Ω × � Ω × � Ω × � r ) Newton’s laws are valid in inertial frames in which the cause of acceleration is a force, whose origin can be traced to a real source. a f and � The cause of the acceleration � A are due to real forces which a f − � give rise to the first term � A . The force is m times this acceleration. Since the linear acceleration of the rotating frame is � A , the acceleration of a particle as seen in the rotating frame is indeed a f − � � A . All other terms in the above cannot be shown to arise from any physical source and hence are termed as Pseudo Force . We drop the third term by assuming � Ω is const ant. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 1 / 10
Centrifugal & Coriolis Forces a f − � A ) − 2 � v r − � Ω × ( � � a r = ( � Ω × � Ω × � r ) The last term − � Ω × ( � Ω × � r ) is the familiar Centrifugal Force . Its direction is radially outward. The second term is the Coriolis Force . This force arises if there is a lateral velocity in the rotating frame itself. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 2 / 10
Example: Effective Gravity on the Earth a f − � A ) − 2 � v r − � Ω × ( � � a r = ( � Ω × � Ω × � r ) Consider an object of mass m on the earth’s surface at a latitude λ . The gravitational force due to the earth is mg vertically downward, i.e., along the line joining the body with the centre of the earth. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 3 / 10
Example: Effective Gravity on the Earth The latitude circle has a radius R cos λ . The centrifugal force m � Ω × ( � r ). The term � Ω × � Ω × � r has a magnitude R ω cos λ and is directed along the latitude circle at the position of the object. The centrifugal force is directed away from the axis and is of magnitude mR ω 2 cos λ . The component of this along the vertical is mR ω 2 cos 2 λ which accounts for the reduction in effective gravity. At the equator the reduction is 0.3%. At any latitude other than at the equator, there is a horizontal component of the centrifugal force having a magnitude m ω 2 R sin λ cos λ . This results in a plumb line not showing true vertical. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 4 / 10
Example 2 : Coriolis Force - A person walking towards the centre of a rotating table a f − � A ) − 2 � v r − � Ω × ( � � a r = ( � Ω × � Ω × � r ) A man is walking from the edge to the centre on a turntable rotating with an angular speed ω in an anticlockwise direction as seen from above. The Coriolis force − 2 m � Ω × � v is directed to his right. In order to be able to walk straight, he will have to counter it with an equal frictional force to his left. How does it arise? Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 5 / 10
Example 2 : Coriolis Force - A person walking towards the centre of a rotating table The torque acting on the person is d � L dt , with | � L | = mr 2 ω Since the person is moving towards the centre with a speed v , dr dt = − v d | � L | = 2 mr dr dt = − 2 mr ω v dt Thus the torque acting on the person is r times ( − 2 m ω v ), which is the tangential friction applied by the man on the turntable. If the man chooses not to apply such a force, in the lab frame he will experience a tangential acceleration. Since there is no horizontal force on the man, in order to keep the angular momentum conserved, as the man tries to reduce his distance to the centre, his tangential speed with respect to turntable has to increase, requiring a tangential accelerattion. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 6 / 10
Example 3: Eastward deflection of a falling body Since the earth is a rotating frame, when a mass is dropped from a height h from the surface of the earth, it will experience a Coriolis force along the horizontal direction. Consider a body being dropped from a height h above the surface of the earth at a latitude λ Take true vertical along the z axis, the direction towards the north as the x-axis and the westward direction as the y-axis. � ω = ω cos λ ˆ x + ω sin λ ˆ z Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 7 / 10
Example 3: Eastward deflection of a falling body z ; � � ω = ω cos λ ˆ x + ω sin λ ˆ v ≈ − gt ˆ z Hence − 2 � ω × � v = − 2 ω gt cos λ ˆ y The Coriolis acceleration is time dependent and eastward. The � t 0 adt = ω g cos λ t 2 eastward velocity is v ( t ) = � Since the time to drop is 2 h / g � √ 2 h / g t 2 dt d = ω g cos λ 0 � = 2 ω h cos λ 2 h 3 g Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 8 / 10
Example 4: Foucault’s Pendulum- Evidence of Earth’s rotation A pendulum is suspended freely from a height of 70m from the great dome of Pantheon in Paris. A needle like projection from the bob touches a bed of sand so that the path of the bob is traced as the pendulum swings. The plane of the motion of the pendulum was found to rotate clockwise, viewed from above, completing one cycle in 32 hours tracing out a trajectory. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 9 / 10
Example 4: Foucault’s Pendulum- Evidence of Earth’s rotation Consider a pendulum on the North Pole. Gravity along true vertical. Gravity and tension defines a plane in which the pendulum swings. The earth rotates below it once in 24 hours. For an inertial observer, the plane of pendulum rotates with a time period of 24 hours. At a latitude λ , the angular velocity vector is � ω = ω cos λ ˆ x + ω sin λ ˆ z The velocity is in the x-y plane. We are only interested in z-component of Coriolis force which acts perpendicular to x-y plane. The magnitude of the force is F = (2 m ω sin λ | v | . This is same as what happens on the north pole with ω being replaced by ω sin λ so that the time period increases by a factor 1 / sin λ . Dipan Ghosh (I.I.T. Bombay) Class. Mech. -34 November 2, 2014 10 / 10
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