The position of a particle of mass 1 kg at time t is given by r = t , i + j + 2t^2 , k , where t is in seconds and the coefficients have the proper units for r to be in metres. What is the component of the angular momentum (with respect to the origin) in kg m ^2 s ⁻¹ along the vector ( i + j ) ?

Step-by-step solution

The position vector of the particle is given by r = t , i + j + 2t^2 , k . The velocity vector is the time derivative of the position vector: v = d r dt = i + 4t , k The linear momentum of the particle is: p = m v = 1 ( i + 4t , k ) = i + 4t , k The angular momentum L with respect to the origin is given by r p : L = (t , i + j + 2t^2 , k ) ( i + 4t , k ) Evaluating the cross product: L = vmatrix i & j & k t & 1 & 2t^2 1 & 0 & 4t vmatrix L = i (1 4t - 0) - j (t 4t - 2t^2 1) + k (0 - 1 1) L = 4t , i - 2t^2 , j - k We