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)`
Prove that `omega_(PO)=2omega_(PA)`

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

The angular velocity of a particle is vec(w) = 4hati + hatj - 2hatk about the origin. If the position vector of the particle is 2hati + 3hatj - 3hatk , then its linear velocity is

The position vector of a particle is r = a sin omega t hati +a cos omega t hatj The velocity of the particle is

For a particle of a rotating rigid body, v = r omega , which of the following are correct ?

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

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

A particle, with restoring force proportional to displacement and resulting force proportional to velocity is subjected to a force F sin omega t . If the amplitude of the particle is maximum for omega = omega_(1) , and the energy of the particle is maximum for omega=omega_(2) , then

A particle, with restoring force proportional to displacement and resulting force proportional to velocity is subjected to a force F sin omega t . If the amplitude of the particle is maximum for omega = omega_(1) , and the energy of the particle is maximum for omega=omega_(2) , then

Position vector of a particle moving in x-y plane at time t is r=a(1- cos omega t)hat(i)+a sin omega t hat(j) . The path of the particle is

the angular velocity omega of a particle varies with time t as omega = 5t^2 + 25 rad/s . the angular acceleration of the particle at t=1 s is