Video Transcript
Three particles move independently
of each other, and each particle is subject to a force perpendicular to the
direction of its instantaneous velocity. What is the total angular momentum
about the origin of the particles? What is the rate of change of the
total angular momentum about the origin of the particles?
We can call the total angular
momentum of the particles capital 𝐿 and the rate of change of that total Δ𝐿 over
Δ𝑡. To start off solving for 𝐿, the
total angular momentum of our three particles 𝑚 one, 𝑚 two, and 𝑚 three about the
origin, we can recall the mathematical definition for angular momentum. 𝐿 as a vector is equal to its
position vector crossed with its momentum. And knowing that 𝑝 momentum is
equal to mass times velocity, we can also write this as 𝑚 times the quantity 𝑟
cross 𝑣.
In our situation, we have three
masses at different locations, and, therefore, each one has its own different
position vector. We can write out what those
position vectors are by observing our mass positions relative to the origin. We see that 𝑚 one is at a position
of negative two meters on the 𝑥-axis and positive one meter on the 𝑦. So, we write its position vector as
negative two 𝑖 plus 𝑗 meters.
The position of mass 𝑚 two
relative to the origin is plus four meters in the 𝑥-direction and plus one meter in
the 𝑦. So, it’s position vector is four 𝑖
plus 𝑗 meters. And 𝑚 three is at a position of
positive two meters in the 𝑥-direction and negative two meters in the 𝑦-direction
relative to the origin. So, we can write its position
vector, 𝑟 three, as two 𝑖 minus two 𝑗 meters.
Since we’re given the velocities
and masses of each one of our three masses, we now have all the information we need
to calculate the angular momentum of each one. Since angular momentum 𝐿 is a
vector and we’ll take a cross product to compute that vector, let’s remind ourselves
of the rule for crossing two different vectors, we’ll call them generically 𝐴 and
𝐵, and assume they’re three-dimensional.
Given two three-dimensional vectors
𝐴 and 𝐵, their cross product is equal to the determinant of the three-by-three
matrix where the first row of the matrix is the unit vectors 𝑖, 𝑗, and 𝑘 of our
three-dimensional space. The last two rows are the 𝑖, 𝑗
and 𝑘 components of these two vectors 𝐴 and 𝐵 respectively. We’ll use this rule to calculate
the angular momentum of mass one, we’ll call it 𝐿 one, and also the same for mass
two, we’ll call it 𝐿 two, and the same thing for mass three. Then, we’ll add together those
three angular momenta to solve for 𝐿, the total.
Starting with 𝐿 one, that’s equal
to 𝑚 one times 𝑟 one cross 𝑣 one. We know 𝑚 one, the mass, and 𝑣
one, the velocity, from our diagram. And we’ve solved for 𝑟 one, the
position vector of that mass relative to the origin. Plugging in for our value 𝑚 one
and following our cross-product rule for setting up the relationship between 𝑟 one
and 𝑣 one, we know that because 𝑟 one and 𝑣 one only have 𝑥- and 𝑦-components,
that means the direction of their cross product will only have a 𝑧-component. Therefore, 𝐿 one will only have a
𝑘-hat direction.
This means that when we evaluate
this matrix, we only need to evaluate the 𝑘th component because the 𝑖- and
𝑗-components will be zero. When we make that evaluation, we
take negative two meters and multiply it by 4.0 meters per second and subtract from
that one meters times zero, which is zero. Overall then, 𝐿 one is 2.0
kilograms times negative 8.0 meters squared per second in the 𝑘-direction, or
negative 16 kilograms meter squared per second in the 𝑘-direction.
Next, we move to calculating 𝐿
two, the same value but for 𝑚 two, 𝑟 two, and 𝑣 two. Again calculating only the
𝑘-component because that’s the only nonzero component to 𝐿 two, we find it’s equal
to 4.0 kilograms times negative 5.0 meters squared per second in the 𝑘-direction,
or negative 20 kilograms meter squared per second 𝑘. Then, we do the same thing for 𝐿
three with 𝑚 three, 𝑟 three, and 𝑣 three. This vector comes out to 6.0
kilograms meter squared per second again in the 𝑘-direction.
Now that we’ve solved for 𝐿 three,
𝐿 two, and 𝐿 one, we’re able to add all these three together and solve for the
overall angular momentum 𝐿. This is equal to negative 16 minus
20 plus 6.0 kilograms meter squared per second in the 𝑘-direction, or negative 30
kilograms meter squared per second in the 𝑘-direction. That’s the total angular momentum
of this system of masses about the origin.
Next, we wanna solve for the time
rate of change of that total angular momentum. We recall from the rotational
version of the impulse-momentum theorem that the change in angular momentum 𝐿 is
equal to the total average torque 𝜏 multiplied by Δ𝑡. This tells us that Δ𝐿 over Δ𝑡 is
equal to the total torque 𝜏. Each one of the three masses
experiences a torque. So, to solve for the overall torque
𝜏, we’ll need to solve for the individual torques and sum them.
Since torque is equal to 𝑟 cross
𝐹, and we’ve already solved for the position vectors of our three masses and are
given their forces, we can use the same cross product rule to solve for the overall
torque. Just like with our angular
momentum, since all of our position vectors and all of the forces acting on our
particles have only 𝑥- and 𝑦-components, that means all of our torques 𝜏 one, 𝜏
two, and 𝜏 three will only have 𝑘-components. And that means we only need to
evaluate the 𝑘-component of our three-by-three matrix.
In the case of 𝜏 one, plugging in
for 𝑟 one and 𝐹 one by component, this is equal to 6.0 newton meters in the
𝑘-direction. Moving onto 𝜏 two, the torque on
particle two, this is equal to the cross product of 𝑟 two and 𝐹 two, or 40 newton
meters in the 𝑘-direction. And lastly, we calculate 𝜏 three,
plugging in for 𝑟 three and 𝐹 three in our matrix. This gives us a result of negative
16 newton meters in the 𝑘-direction.
With values now for 𝜏 three, 𝜏
two, and 𝜏 one, we can add these together to solve for the overall torque 𝜏. This is equal to 6.0 plus 40 minus
16 newton meters in the 𝑘-hat direction, or 30 newton meters in the 𝑘-direction. And since this is the overall
torque acting on our system of masses, by the impulse-momentum theorem, it’s also
equal to the time rate of change of angular momentum.
