Question

An object with mass m=3

is moving along the trajectory r⃗ (t)=2i^+t2j^+(t−t3)k^
. The x
-component of the object's angular momentum about the origin at time t=5
will be about

Answer 1

`(-1950)`.

Step-by-step explanation:

At time
`t`, the angular momentum
`\vec{L}(t)` of the object with respect to the origin is equal to the vector cross-product of the position vector
`\vec{r}(t)` and the momentum vector
`\vec{p}(t)`.

In this question, only the position vector as a function of time,
`\vec{r}(t)`, is given. The momentum vector can be found by multiplying the mass of the object by the velocity vector
`\vec{v}(t)`. The velocity vector isn't directly given but can be found by differentiating position
`\vec{r}{(t)\!` with respect to time
`t`.

To find the velocity vector
`\vec{v}(t)`, differentiate the position vector
`\vec{r}(t)` with respect to time
`t`:

`\begin{aligned}\vec{v}(t) &= (d)/(d t)\left[\vec{r}(t)\right] \\ &= (d)/(d t) \begin{bmatrix}2 \\ t^(2) \\ t - t^(3)\end{bmatrix} \\ &= \begin{bmatrix}0 \\ 2\, t \\ 1 - 3\, t^(2)\end{bmatrix}\end{aligned}`.

Multiply the velocity vector by the mass of the object
`m = 3` (a scalar) to find the momentum vector
`\vec{p}(t)`:

`\begin{aligned}\vec{p}(t) &= m\, \vec{v}(t) \\ &= 3\, \begin{bmatrix}0 \\ 2\, t \\ 1- 3\, t^(2)\end{bmatrix} \\ &= \begin{bmatrix}0 \\ 6\, t \\ 3 - 9\, t^(2)\end{bmatrix}\end{aligned}`.

The angular momentum of this object with respect to the origin is equal to the vector cross-product of position with momentum:
`\begin{aligned}\vec{p}(t) &= \vec{r}(t) * \vec{p}(t)\end{aligned}`. Note that vector cross-products are not commutative, and the position vector
`\vec{r}(t)` should be placed to the left of the momentum vector
`\vec{p}(t)`.

`\begin{aligned}\vec{p}(t) &= \vec{r}(t) * \vec{p}(t) \\ &= \begin{bmatrix}2 \\ t^(2) \\ t - t^(3)\end{bmatrix}* \begin{bmatrix}0 \\ 6\, t \\ 3 - 9\, t^(2)\end{bmatrix} \\ &= \begin{bmatrix}t^(2)\, (3 - 9\, t) - (t - t^(3))\, (6\, t)) \\ 2\, (3 - 9\, t^(2)) - (t - t^(3))\, (0) \\ t^(2)\, (0) - 2\, (6\, t)\end{bmatrix} \\ &= \begin{bmatrix}-3\, t^(2) - 3\, t^(4) \\ 6 - 18\, t^(2) \\ -12\, t\end{bmatrix}\end{aligned}`.

Substitute in
`t = 5` and evaluate to obtain the
`x`-component of this vector:

`-3\, (5)^(2) - 3\, (5)^(4) = -75 - 1875 = (-1950)`.

In other words, the
`x`-component of the angular momentum about the origin would be
`(-1950)` at
`t = 5`.