### Video Transcript

A particle of mass 5.0 kilograms has position vector π« equals 2.0π’ minus 3.0π£ meters at a particular instant of time π‘. At the instant π‘, the velocity of the particle with respect to the origin is π― equals 3.0π’ meters per second. What is the angular momentum of the particle at the instant π‘? If a force π
equals 5.0π£ newtons acts on the particle at the instant π‘, what is the torque about the origin in that instant?

Letβs highlight the vital
information given in this two-part problem. Weβre told that the mass of the
particle weβre considering is 5.0 kilograms and that it has a position vector π«
equals 2.0π’ minus 3.0π£ meters. The particle is moving with a
velocity of π£ equals 3.0π’ meters per second. And in part two, weβre told that
the particle was subjected to a force π
equals 5.0π£ newtons.

In part one, we want to solve for
the angular momentum of the particle at time π‘; weβll call that capital π. And in part two, weβre asked to
solve for the torque about the origin; weβll represent that using the Greek letter
π.

Both angular momentum and π, the
torque, are vectors, meaning they have both magnitude and direction. Letβs begin our solution by
summarizing the information given and recalling the vector relationship for angular
momentum.

As we work towards solving for the
angular momentum of the particle, letβs recall the relationship between π« and π― for angular momentum. The angular momentum of an object
is equal to the position vector of that object crossed with the objectβs
momentum.

We know we can write momentum π as
the objectβs mass times its velocity, so angular momentum equals π« cross π times
π―. As we apply this equation to our
scenario, we can factor out the π so that the equation reads π equals π times the
quantity π« cross π―.

Now it will be important to recall
mechanically how do we compute the cross-product between two vectors. In this case, we have π« and π―
given to us as part of the problem statement. To know how to combine these in the
cross-product, letβs recall the general equation for the cross-product between two
three-dimensional vectors π and π.

The cross-product between two
vectors is equal to their components combined according to this equation. Notice that the resulting vector
has an π’-, a π£-, and a π€-component and that each one of these components is
comprised of the components of π and π that are perpendicular to that
direction. For example, the π’th component of
π cross π involves the π¦- and π§-components of π and π, respectively.

Letβs apply this cross-product
equation to the given information in this problem. First, letβs figure out the π’th
component of the angular momentum π.

When we write out the π’-, π£-, and
π€-components of the angular momentum, we see, as we look at the first π’th
component, that that component is equal to π sub π¦ times π£ sub π§ minus π sub π§
times π£ sub π¦.

But for the position and velocity
vectors given, there is no π§-component, so π£ sub π§ and π sub π§ are both
zero. We have both those terms being
zero; the entire π’th component is zero as well.

So now we move on to the π£th
component of our angular momentum. Again, we see π sub π§ in the
first term and π£ sub π§ in the second. Since those are both still zero,
the π£th component of π is zero too.

Now on to the last component, the
π€-direction component: π sub π₯ minus π£ sub π¦ minus π sub π¦ times π£ sub
π₯. Looking at our velocity vector, it
has only an π- or π₯-direction component. That means π£ sub π¦ is zero.

π sub π¦ is negative 3.0 and π£
sub π₯ is 3.0. So our π€-component simplifies to
negative negative 3.0 times 3.0, or 9.0. We can now rewrite our expression
for angular momentum in a simplified way. π, the angular momentum of our
system, is equal to the mass π multiplied by 9.0 meters squared per second in the
π€-direction.

Weβre given π as 5.0
kilograms. And when we multiply these two
numbers together, we find that the overall angular momentum π, the product of these
two numbers, is 45 kilograms meters squared per second in the π€-direction. Thatβs the magnitude and direction
of the systemβs angular momentum.

In part two of our problem, weβre
told that a force π
equals 5.0 newtons in the π£-direction acts on the
particle. And we want to solve for the torque
this creates.

Letβs recall the vector
relationship for torque. The torque that acts on an object
is equal to the position vector of that object cross the force. In our problem, we have both π« and
π
, and we need to take a cross-product between them to solve for torque π.

Letβs write out the general
equation for the cross-product between π« and π
using our cross-product
formula. With the components of torque
written out, we can now evaluate the π’-, π£-, and π€-components of π to see what
their value is.

Looking at the π’th component, we
see that involves πΉ sub π§ in the first term and π sub π§ in the second term. Looking at our π« and π
vectors,
we see that neither of them has a π§-component, so πΉ sub π§ and π sub π§ are both
zero. That means that the π’-component of
π is zero as well.

Moving on to the π£-component, we
again see π sub π§ in the first term and πΉ sub π§ now in the second term. These values are zero and so so is
the π£th component of π.

Moving to the last component, the
π€th component, π sub π₯ is 2.0, πΉ sub π¦ is 5.0, π sub π¦ is negative 3.0, and
πΉ sub π₯ is zero. This means the π€th component of π
can be written as 2.0 times 5.0.

When we multiply these numbers
together, we get 10. So this tells us the overall torque
acting on our particle is simply 10 newton meters acting in the π€-direction. Weβve now solved for both the
angular momentum and the torque that acts on this particle.