Friday, May 6, 2011

Introduction to Basic Physics (Mechanics) Pt. 4

Energy Analytics for a Pendulum:

We now examine in more detail the mechanics of the simple pendulum, with a view to the energy constraints and analytics. We make use of the diagram shown above. We note that the difference between the starting height for the initiation of the oscillation and midpoint height (or the minimum height of mass m at the middle of the oscillation) is delta y. Ideally, the loss in potential energy equals the gain in kinetic energy or:

mg(delta y) = ½ m v^2

Then the velocity at the point at height delta y is:
v = [2g (delta y)]^½

This can also be developed in terms of the angle Θ.

Let delta y = vertical displacement of the mass upwards. Then, from the geometry of the situation one finds (we use ℓ for the length to avoid confusion with the Lagrangian L):

delta y = ℓ - ℓ cos Θ

And the change in potential energy of the system is:

mg delta y = mg ℓ (1 - cos Θ)

It is also possible to work out a key quantity called “the Lagrangian” which has supreme importance in mechanics, as the difference between kinetic (T) and potential (V) energies, hence:

L = T – V

Example Problem:

A 0.1 kg mass m attached to a length of pendulum string ℓ = 0.50 m swings through an angle Θ = 30 degrees to reach its minimum position at the center of its oscillation. Assuming g = 10 m/s/s, find:

a) The maximum velocity attained

b) The maximum kinetic energy attained.

c)The potential energy lost to attain (b) assuming no dissipative forces or less of energy.

d)The Lagrangian of the System


1)the maximum velocity is given by:

v = [2g (delta y)]^½

= [2g delta y]^½ = [2g (ℓ (1 - cos Θ))]^½

[(2(10m/s^2)(0.5m(1 – cos(30))]^½

= [20 {0.066}]^½ = 1.15 m/s

b) Maximum KE attained = T(max) = ½ m v^2

= ½ (0.01kg)(1.15 m/s)^2

T(max) = 0.006 Joule

c) The potential energy lost to obtain T is 0.006 J (assume conservation of mechanical energy, so potential energy lost in system equals kinetic energy gained)

d) The Lagrangian of the system is defined: L = T – V

with V = mg (delta y) = mgℓ(1 - cos Θ)

and T = ½ m v^2

Then: L = ½ m v^2 - mg ℓ(1 - cos Θ)

= m[(v^2 /2 - gℓ(1 - cos Θ)]

Next: Orbits

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