Angular Position

The angular acceleration of a wheel is alpha=6.0t^(4)-4.0t^(2) , with alpha in radians per second-squared and t in seconds. At time t = 0, the wheel has an angular velocity of +2.5 rad/s and an angular position of +1.5 rad. Write expressions for (a) the angular velocity (rad/s) and (b) the angular position (rad) as functions of time (s).

Light of wavelength 500 nm falls normally on a slit of width 1 mu m producing Fraunhofer diffraction pattern on a screen. Calculate the angular position of the first minimum and the angular width of the central maximum.

In Young's double slit experiment, distance between two sources is 0.1mm. The distance of screen from the sources is 20cm. Wavelength of light used is 5460 Å . Then, angular position of first dark fringe is approximately

In young's double slit experiment the distance between two sources is 0.1//pimm .the distance of the screen from the source is 25cm .wavelength of light used is 5000Å .Then the angular position of the first dark fringe is-

The small sphericla balls are free to move on the inner surface of the rotating spherical chamber of radius R=0.2 m If the balls reach a steady at angular position theta=45^(@) , the anglular speed omega of devices is

A coin is pushed down tangentially from an angular position theta on a cylindrical surface , with a velocity v as shown in figure If the coefficient of friction the coin and surface is mu find the tangential acceleration of the coin

Two coherent sources separated by distance d are radiating in phase having wavelength lambda . A detector moves in a big circle around the two sources in the plane of the two sources. The angular position of n=4 interference maxima is given as

In Young's double slit experiment, the distnace between two sources is 0.1//pimm . The distance of the screen from the source is 25 cm. Wavelength of light used is 5000Å Then what is the angular position of the first dark fringe.?