At time t=0, a 2kg particle has position vector r=(5.0m)i+(-8.0m)j relative to the origin. Its velocity just then is given by v=(-5.0t^2m/s)i. For the following answers, use t for the time. A) About the origin and for t>0, what is the particle's angular momentum? B) About the origin and for t>0, what is the torque acting on the particle? C) Repeat A and B for a point with coordinates (-7.0m, -4.0m, 0.0m) instead of the origin.

Please don't post under different names.

I will be happy to critique your thinking on this.

To calculate the particle's angular momentum about a certain point, we need to know both the particle's velocity and its position relative to that point. The angular momentum L is defined as the cross product of the position vector r and the linear momentum p:

L = r x p

Let's solve these problems step by step:

A) To find the particle's angular momentum about the origin, we need to calculate the cross product of the position vector and the linear momentum. Given that the position vector r = (5.0m)i + (-8.0m)j and the linear momentum p = (-5.0t^2m/s)i, we can calculate the cross product as follows:

L = r x p
= (5.0m)i x (-5.0t^2m/s)i
= 5.0 * (-5.0t^2)m^2/s * k

Therefore, the particle's angular momentum about the origin is given by L = -25t^2 m^2/s * k (where k is the unit vector in the z-direction).

B) To find the torque acting on the particle about the origin, we need to calculate the time derivative of the angular momentum. Differentiating the above expression for angular momentum with respect to time t, we get:

dL/dt = d/dt(-25t^2 m^2/s * k)
= -50t m^2/s^2 * k

So the torque acting on the particle about the origin is given by τ = -50t m^2/s^2 * k.

C) To repeat parts A and B for a point with coordinates (-7.0m, -4.0m, 0.0m), we can apply the same logic. We calculate the position vector from this point to the particle's position at time t=0, and then proceed to find the angular momentum and torque.

The position vector r' from the new point to the particle's position is:

r' = r - r0
= (5.0m)i + (-8.0m)j - (-7.0m)i - (-4.0m)j
= 12.0m*i - 4.0m*j

Now, using this new position vector r' and the linear momentum p = (-5.0t^2m/s)i, we can calculate the particle's angular momentum about the new point as follows:

L' = r' x p
= (12.0m*i - 4.0m*j) x (-5.0t^2m/s)i

Taking the cross product of these vectors, we find:

L' = 60t^2 m^2/s * k

Therefore, the particle's angular momentum about the point (-7.0m, -4.0m, 0.0m) is given by L' = 60t^2 m^2/s * k.

Similarly, to find the torque about the new point, we differentiate the above expression for angular momentum with respect to time t:

dL'/dt = d/dt(60t^2 m^2/s * k)
= 120t m^2/s^2 * k

So the torque acting on the particle about the point (-7.0m, -4.0m, 0.0m) is given by τ' = 120t m^2/s^2 * k.

L = R x m V

where R is position vector from reference point origin for a and b) to mass
V is velocity vector
x means cross or vector product

the x coordinate of the position vector is 5 at the start but changes with time because the thing has x velocity of -5t^2
integral = 5-(5/3) t^3

so for part a
L = (5-5t^3/3) i - 8 j)x(-5 t^2 i + 0 j)

(5-5t^3/3) -8 0
-5t^2 0 0 = -40 t^2 k
i j k

for part b
Torque = rate of change of angular momentum
dL/dt = d/dt(40 t^2 k) = 80 t

for part c
Rx = 5-5t^3/3 - (-7) = 12 - 5 t^3/3
Ry = -8 -(-4) = -4
Rz = 0 still
just do it again

(5-5t^3/3) -8 0

-5t^2 0 0 = +40 t^2 k
i j k