Lecture On Theoretical Mechanics - 3

PBC Lecture Notes Series in Physics Lecture 2 Prepared by Dr. Abhijit Kar Gupta, e-mail: [email protected]

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Motion a Particle-II We calculate the velocity, acceleration in different coordinate systems. Once again we emphasize that the position vector is very important and this is our starting point. In Spherical Polar Coordinate system (r , θ , ϕ ): Position vector, r = iˆx + ˆjy + kˆz

= iˆ.r sin θ . cos ϕ + ˆj.r sin θ . sin ϕ + kˆ.r cos θ

The basis vectors, eˆr =

∂r ∂r

∂r = iˆ sin θ cos ϕ + ˆj sin θ sin ϕ + kˆ cos θ , ∂r

eˆθ =

∂r ∂θ

∂r = iˆ cos θ cos ϕ + ˆj cos θ sin ϕ − kˆ sin θ , ∂θ

eˆϕ =

∂r ∂ϕ

∂r = − iˆ sin ϕ + ˆj cos ϕ . ∂ϕ

When we consider the motion, we should remember that the unit vectors eˆr , eˆθ , eˆϕ are also changing with time. They change as the directions θ and ϕ change. The time derivatives, •

eˆr =

• • • • • deˆr = (cos θ cos ϕ θ − sin θ sin ϕ ϕ ).iˆ + (cos θ sin ϕ θ − sin θ cos ϕ ϕ ). ˆj − sin θ .θ .kˆ dt •

•

= (cos θ cos ϕ .iˆ + cos θ sin ϕ . ˆj − sin θ .kˆ).θ + sin θ (− sin ϕ .iˆ + cos ϕ . ˆj ) ϕ [Rearranging the terms] •

•

•

∴ eˆr = θ eˆθ + sin θ ϕ eˆϕ

(1)

Similarly, • • • deˆ eˆθ = θ = − θ eˆr + cos θ ϕ eˆϕ and dt • • • deˆϕ eˆϕ = = − sin θ ϕ eˆr − cos θ ϕ eˆθ . dt

(2) (3)

Now the position vector can be written as •

•

•

•

•

r = reˆr •

∴ Velocity, V = r = r eˆr + r eˆr = r eˆr + r θ eˆθ + r. sin θ ϕ . eˆϕ .

PBC Lecture Notes Series in Physics Lecture 2 Prepared by Dr. Abhijit Kar Gupta, e-mail: [email protected]

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Acceleration, a = V = r ••

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• •

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• •

= r eˆr + r eˆr + r θ eˆθ + r θ eˆθ + r θ eˆθ + r . sin θ ϕ . eˆϕ + r. cos θ θ ϕ . eˆϕ + ••

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r. sin θ ϕ . eˆϕ + r. sin θ ϕ . eˆϕ ••

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• •

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= r eˆr + r ( θ eˆθ + sin θ ϕ eˆϕ ) + r θ eˆθ + r θ eˆθ + r θ ( − θ eˆr + cos θ ϕ eˆϕ ) + r . sin θ ϕ . eˆϕ + • •

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r. cos θ θ ϕ . eˆϕ + r. sin θ ϕ . eˆϕ + r. sin θ ϕ . ( − sin θ ϕ eˆr − cos θ ϕ eˆθ )

[ using (1), (2) and (3)] • 2

••

• 2

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• 2

• •

∴ a = ( r − r θ − r sin 2 θ ϕ ).eˆr + ( r θ + 2 r θ − r. sin θ cos θ ϕ . ) eˆθ + ••

• •

• •

( r sin θ ϕ + 2 r ϕ sin θ + 2r cos θ θ ϕ ) eˆϕ ………………..(4) •

•

When the motion is such that the angles θ and ϕ vary at constant rate, θ = ϕ = const ••

••

⇒ θ = ϕ = 0 , we have the following expression: • 2

••

• 2

• 2

• •

• •

∴ a = ( r − r θ − r sin 2 θ ϕ )eˆr + ( 2 r θ − r sin θ cos θ ϕ . ) eˆθ + ( 2 r ϕ sin θ + • •

2r cos θ θ ϕ ) eˆϕ . ……………….(5) •

•

Further when θ and ϕ remain fixed, we can write θ = ϕ = 0 and we get back the trivial case, ••

a = r eˆr ,

only translational motion (no rotation).

Motion on a sphere: Consider the motion of the particle on a sphere, then we can write •

••

r = const ⇒ r = r = 0 . Thus from (4) we can write, • 2

• 2

• 2

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• •

a = ( − r θ − r sin 2 θ ϕ ).eˆr + ( r θ − r sin θ cos θ ϕ . ) eˆθ + ( r sin θ ϕ + 2r cos θ θ ϕ ) eˆϕ . Motion on a plane: •

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ϕ = const. ⇒ ϕ = ϕ = 0 From (4), we can write, ••

• 2

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• •

a = ( r − r θ )eˆr + ( r θ + 2 r θ ) eˆθ . [This is the same result that we obtained with plane polar coordinate system.]

PBC Lecture Notes Series in Physics Lecture 2 Prepared by Dr. Abhijit Kar Gupta, e-mail: [email protected]

3

Motion on a circle: For the motion of the particle on a circle, we can further consider r = const. • 2

••

∴ a = − r θ eˆr + r θ eˆθ .

If the force on the particle of mass m is given by F = F (r )eˆr , we can write from Newton’s 2nd law of motion: • 2

••

F (r )eˆr = m a = m ( − r θ eˆr + r θ eˆθ ). The corresponding scalar equations, •2

− mr θ = F (r )

(i)

••

mr θ = 0

(ii) • 1d (mr 2 θ ) = 0, keeping in mind that r = const. We can manipulate eqn. (ii) and write, r dt •

mr 2 θ = const.

∴

Thus we arrive at something constant for the motion. Now what is this? To know, we consider the following cross product of two vectors, •

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r × m r = reˆr × mr θ eˆθ

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[Q r = r eˆr + r θ eˆθ in a plane. For circle, r = r θ eˆθ . ]

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= mr 2 θ nˆ , where nˆ is a vector perpendicular to the plane of eˆr and eˆθ . The above is moment of momentum which is nothing but angular momentum. •

Thus we can say that the angular momentum, mr 2 θ is constant for the kind of force we choose. This is a special case; later we shall know in a more general way that this kind of motion is called central force motion, applicable for planetary orbits. Exercise:

Find the expressions for velocity and acceleration in cylindrical coordinate system ( ρ , ϕ , z ) for the motion of a particle. =======================