A thin rod of length 0.75 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 2.76 rad/s. Neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

a) KE = 0.0857Joules

b) height = 0.073m

Explanation:

The rotational inertia of the rod at an axis passing through the fixed end is given by:

I = Icm + my

I= 1/12 mL^2 + m(L/2)

I= 1/12 ×0.12×(0.75)^2 + 0.12× (0.75/2)^2

I = 5.625×10^-3 + 0.01688

I = 0.0225Kg/m^-2

a) KE = 1/2 Iw^2

Where I= rotational inertia

w= angular speed at lowest point

KE = 1/2 × 0.0225 ×(2.76)^2

KE= 0.0857 Joules

b) Ki + Ui = Kf + Uf

Ki = Uf - Ui = 0

mgh = K

0.0857= mgh

h = 0.0857/( 0.120× 9.8)

h = 0.0857/ 1.176

h = 0.073m

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.