Calculating Angular Momentum with Constant Torque: A Particle in Motion

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lebprince
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Homework Statement



At t = 15 s, a particle has angular momentum <5, 8, -3> kg · m2/s relative to location A. A constant torque <12, -13, 20> N · m relative to location A acts on the particle. At t = 15.2 s, what is the angular momentum of the particle relative to location A?


The Attempt at a Solution



i know that torque = R x Fnet , but am not sure how to use it in this problem here. any help please?
 
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Isn't the torque equal to the first derivative of the angular momentum?

So wouldn't that mean that the rate of change of the angular momentum is expressed by the Torque given relative to point A?
 
LowlyPion said:
Isn't the torque equal to the first derivative of the angular momentum?

So wouldn't that mean that the rate of change of the angular momentum is expressed by the Torque given relative to point A?


i still don't get how to solve the problem
 
rl.bhat said:
Torque =( L2-L1)/t
Torque*t = L2 - L1.
Find L2.


oh u meant deltat thanks
 
Yes. Like that.

The Δt in this case is .2 s and is a scalar to the Torque Vector which as rl.bhat indicated should yield a difference vector ΔLa that added to the original La vector should yield the New La at T + Δt.