A particle moving along a circule path of radius 'r' with uniform angular velocity `omega`. Its angular acceleration is

A particle is moving along a circular path of radius 'r' with uniform angular velocity omega . Its angular acceleration is

A particle moving in a circular path of certain radius, with uniform angular velocity omega has an angular acceleration equal to :

A particles is moving on circular path of radius R with constant angular velocity omega . Its centripetal acceleration will be …... .

A particle of mass m is revolving along a circular path of radius r with uniform angular velocity omega . What will be the magnitude and direction of the angular momentum of that particle ?

A particle is moving on a circular path of radius R with uniform angular speed omega . The magnitude of average velocityof particle during time t=0 to t=(2pi)/(3 omega) :

A particle is moving in a circle of radius r with constant angular velocity omega . At any point (r, theta ) on its path , its position vector is vec(r ) = r cos theta hat(i) + r sin theta hat(j) . Show that the velocity of the particle has no component along the radius.

A particle is moving in a circular orbit of radius r_(1) with an angular velocity omega_(1) It jumps to another circular orbit of radius r_(2) and attains an angular velocity omega_(2) . If r_(2)= 0.5 r_(1) , and if no external torque is applied to the system, then the new angular velocity omega_(2) is given by

A particle is moving in a circular orbit of radius r_(1) with an angular velocity omega_(1) . It jumps to another circular orbit of radius r_(2) and attains an angular velocity omega_(2) . IF r_(2) = 0.5 r_(1) and assuming that no external torque is applied to that system, then, the angular velocity omega_(2) is given by