(a) The angular momentum of the particle about the origin is −35.0î − 10.0ĵ + 25.0k kg · m²/s.
(b) The torque on the particle about the origin is −171.0î + 10.0ĵ + 99.0k N · m.
(c) The time rate of change of the particle's angular momentum about the origin at this instant is 0 kg · m²/s².
(a) To find the angular momentum of the particle about the origin, we use the formula L = r x p, where L is the angular momentum, r is the position vector, and p is the momentum vector. In this case, the position vector is given as r = (4.0î − 5.0ĵ + 6.0k) m.
To calculate the momentum vector, we multiply the mass of the particle (given as 1.0 kg) by its velocity vector v = (−1.0î + 5.0ĵ + 1.0k) m/s. Thus, p = 1.0 kg * (−1.0î + 5.0ĵ + 1.0k) m/s.
Taking the cross product of r and p, we have:
L = r x p
L = (4.0î − 5.0ĵ + 6.0k) m x (−1.0î + 5.0ĵ + 1.0k) m/s
Using the properties of the cross product, we can calculate L:
L = [(−5.0)(1.0) − (6.0)(5.0)]î + [(6.0)(−1.0) − (4.0)(1.0)]ĵ + [(4.0)(5.0) − (−5.0)(−1.0)]k
L = −35.0î − 10.0ĵ + 25.0k
Therefore, the angular momentum of the particle about the origin is −35.0î − 10.0ĵ + 25.0k kg · m²/s.
(b) The torque on the particle about the origin can be determined by taking the cross product of the position vector r and the force vector F = (11.0î + 16.0ĵ) N. Using the formula τ = r x F, we have:
τ = (4.0î − 5.0ĵ + 6.0k) m x (11.0î + 16.0ĵ) N
Applying the properties of the cross product, we obtain:
τ = [(−5.0)(16.0) − (6.0)(11.0)]î + [(6.0)(11.0) − (4.0)(16.0)]ĵ + [(4.0)(16.0) − (−5.0)(11.0)]k
τ = −171.0î + 10.0ĵ + 99.0k
Hence, the torque on the particle about the origin is −171.0î + 10.0ĵ + 99.0k N · m.
(c) The time rate of change of the particle's angular momentum about the origin at this instant, denoted as dL/dt, can be obtained by taking the derivative of the angular momentum expression with respect to time.
Differentiating the expression:
dL/dt = d/dt [−35.0î − 10.0ĵ + 25.0k] kg · m²/s
Since there are no time-dependent quantities in the angular momentum expression, the derivative with respect to time is zero.
Therefore, the time rate of change of the particle's angular momentum about the origin at this instant is 0 kg · m²/s².