A particle of mass m moves in the xy -plane such that its position vector at time t is given by r = R( t i + t j ) , where R and are positive constants. If v and a represent the velocity and acceleration vectors of the particle respectively, and L represents its angular momentum about the origin, then the direction of the vector ( v a ) is

Step-by-step solution

The position vector is given by r = R( t i + t j ) . Differentiating with respect to time to find velocity: v = d r dt = R(- t i + t j ) Differentiating velocity to find acceleration: a = d v dt = - ^2 R( t i + t j ) = - ^2 r Now, compute the cross product ( v a ) : v a = [ R(- t i + t j )] [- ^2 R( t i + t j )] v a = - ^3 R^2 [(- t)( t)( i j ) + ( t)( t)( j i )] v a = - ^3 R^2 [- ^2 t k - ^2 t k ] = ^3 R^2 k The angular momentum L about the origin is: L = r m v = m [R( t i + t j )] [ R(- t i + t j )] L = m R^2 ( ^