Angular velocity of a particle is given w.r.t. the point which the position vector of particle is considered.
Here `omega_(PO)=(dalpha)/(dt)`
`omega_(PA)=(dbeta)/(dt)`
`omega_(PO)=2omega_(PA)`

The angular velocity of a particle is given by omega=1.5t-3t^(2)+2 , Find the time when its angular acceleration becomes zero.

A particle moves so th at its p o sitio n vector is omega Where is a constant which of the following is true

The displacement of a particle varies with time according to the relation y= a sin omega t+b cos omega t ……………

The displacement of particle along the X-axis is given by x= a sin^(2) omega t . The motion of the particle corresponds to………..

The displacement of a particle is represented by the equation y= sin^(3) omega t . The motion is………..

The displacement of a particle represented by the equation y= 3 cos((pi)/(4)-2 omega t) . The motion of the particle is

A disc is rotating with angular velocity omega . A force F acts at a point whose position vector with respect to the axis of rotation is r. The power associated with torque due to the force is given by

Path traced by a moving particle in space is called trajectory of the particle. Shape of trajectiry is decided by the forces acting on the particle. When a coordinate system is associated with a particle motion, the curve equation in which the particle moves [y=f(x)] is called equation of trajectory. It is just giving us the relation among x and y coordinates of the particle i.e. the locus of particle. To find equation of trajectory of a particle, find first x and y coordinates of the particle as a function of time eliminate the time factor. The position vector of car w.r.t. its starting point is given as vec(r)=at hat(i)- bt^(2) hat(j) where a and b are positive constants. The locus of a particle is:-

What is the linear velocity of a particle of a uniform circular motionk of radius A and angular velocity omega ?