" (b) "theta=omega_(0)t+(1)/(2)alpha t^(2)

Derive the three equation of rotational motion (i) omega = omega_(0) + at (ii) theta = omega_(0)t + 1/2alpha t^(2) (iii) omega^(2) = omega_(0)^(2) + 2alpha theta Under constant angular acceleration. Here symbols have usual meaning.

The angular displacement of a particle is given by theta=omega_(0)t+(1)/(2) alpha^(2), "and" alpha are constatnt velocity at time, t=2 sec will be ("in" rad//sec)-

Drive the equations of rotatory motion, omega^2 - omega_0^2 = 2 alpha theta and theta= omega_0^t + 1/2 alpha t^2 , where every letter has its usual meaning.

Derive the following relations by calculus method : theta=omega_@t+1/2alphat^2

Obtain the equation omega = omega_(0) alpha t from first principles.

The range of real constant t for which (1-tan^(2)t)sin theta^(2)+tan^(2)t*tan theta^(2)>=theta^(2) always holds AA theta in(0,(pi)/(2)) is [alpha,beta) then (beta)/(alpha) is equal to

Find the dimensions of a. angular speed omega angular acceleration alpha torque tau and d. moment of interia I . . Some of the equations involving these quantities are omega=(theta_2-theta_1)/(t_2-t_1), alpha = (omega_2-omega_1)/(t_2-t_1), tau= F.r and I=mr^2 The symbols have standard meanings.

Find the dimensions of a. angular speed omega angular acceleration alpha torque tau and d. moment of interia I. . Some of the equations involving these quantities are omega=(theta_2-theta_1)/(t_2-t_1), alpha = (omega_2-omega_1)/(t_2-t_1), tau= F.r and I=mr^2 The symbols have standard meanings.

Find the dimensions of a. angular speed omega angular acceleration alpha torque tau and d. moment of interia I . . Some of the equations involving these quantities are omega=(theta_2-theta_1)/(t_2-t_1), alpha = (omega_2-omega_1)/(t_2-t_1), tau= F.r and I=mr^2 The symbols have standard meanings.