A particle of mass 2 kg moves such that its position vector as a function of time t is given by r (t) = 2t^2 i + 3t j + k , where r is in meters and t is in seconds. The torque acting on the particle about the origin at t = 2 s (in N m ) is:

Step-by-step solution

The position vector of the particle is r (t) = 2t^2 i + 3t j + k . Velocity is the first derivative of position with respect to time: v (t) = d r dt = 4t i + 3 j Acceleration is the derivative of velocity: a (t) = d v dt = 4 i From Newton's Second Law, the force acting on the particle is: F = m a = 2(4 i ) = 8 i N At t = 2 s , the position vector is: r (2) = 2(2)^2 i + 3(2) j + k = 8 i + 6 j + k m The torque about the origin is: = r F = (8 i + 6 j + k ) (8 i ) = 8( i i ) + 48( j i ) + 8( k i ) Since i i = 0 , j i =