A particle of mass 0.01 kg having position vector `vecr = ( 10 hati + 6 hatj)` meters is moving with a velocity `5 hati `m/s . Calculate its angular momentum about the origin.

A unit mass at position vector vecr=(3hati+hatj) is moving with velocity vecv=(5hati-6hatj) . What is the angular momentum of the body about the origin?

A partical has the position vector r = hati - 2 hatj + hatk and the linear momentum p = 2 hati - hatj + hatk its angular momentum about the origin is

If postion vector of a particle is given by vec(r) = 10 alpha t^2 hati +[5 beta t- 5]hatj . Find time when its angular momentum about origin is O.

If angular velocity of a particle having position vector vec r = (2hati - 2hatj + 2hatk)m about the origin is vec w = (2hati - 2hatj - hatk)rad/s then magnitude of linear velocity of the particle will be

If angular velocity of a particle having position vector vec r = (2hati - 2hatj + 2hatk)m about the origin is vec w = (2hati - 2hatj - hatk)rad/s then magnitude of linear velocity of the particle will be

A particle of mass 2 kg located at the position (hati + hatj)m has velocity 2(hati - hatj + hatk) m//s . Its angular momentum about Z-axis in kg m^(2)//s is

A particle of mass 2kg is moving such that at time t its position in meter is given by barr(t)=5hati-2t^(2)hatj . The angular momentum of the particle at t=2s about the origin in kgm^(-2)s^(-1) is