A particle of mass 5.0 kilograms has position vector 𝐫 equals 2.0𝐢 minus 3.0𝐣 meters at a particular instant of time 𝑡. At the instant 𝑡, the velocity of the particle with respect to the origin is 𝐯 equals 3.0𝐢 meters per second. What is the angular momentum of the particle at the instant 𝑡? If a force 𝐅 equals 5.0𝐣 newtons acts on the particle at the instant 𝑡, what is the torque about the origin in that instant?
Let’s highlight the vital
information given in this two-part problem. We’re told that the mass of the
particle we’re considering is 5.0 kilograms and that it has a position vector 𝐫
equals 2.0𝐢 minus 3.0𝐣 meters. The particle is moving with a
velocity of 𝑣 equals 3.0𝐢 meters per second. And in part two, we’re told that
the particle was subjected to a force 𝐅 equals 5.0𝐣 newtons.
In part one, we want to solve for
the angular momentum of the particle at time 𝑡; we’ll call that capital 𝐋. And in part two, we’re asked to
solve for the torque about the origin; we’ll represent that using the Greek letter
Both angular momentum and 𝛕, the
torque, are vectors, meaning they have both magnitude and direction. Let’s begin our solution by
summarizing the information given and recalling the vector relationship for angular
As we work towards solving for the
angular momentum of the particle, let’s recall the relationship between 𝐫 and 𝐯 for angular momentum. The angular momentum of an object
is equal to the position vector of that object crossed with the object’s
We know we can write momentum 𝑝 as
the object’s mass times its velocity, so angular momentum equals 𝐫 cross 𝑚 times
𝐯. As we apply this equation to our
scenario, we can factor out the 𝑚 so that the equation reads 𝐋 equals 𝑚 times the
quantity 𝐫 cross 𝐯.
Now it will be important to recall
mechanically how do we compute the cross-product between two vectors. In this case, we have 𝐫 and 𝐯
given to us as part of the problem statement. To know how to combine these in the
cross-product, let’s recall the general equation for the cross-product between two
three-dimensional vectors 𝐀 and 𝐁.
The cross-product between two
vectors is equal to their components combined according to this equation. Notice that the resulting vector
has an 𝐢-, a 𝐣-, and a 𝐤-component and that each one of these components is
comprised of the components of 𝐀 and 𝐁 that are perpendicular to that
direction. For example, the 𝐢th component of
𝐀 cross 𝐁 involves the 𝑦- and 𝑧-components of 𝐀 and 𝐁, respectively.
Let’s apply this cross-product
equation to the given information in this problem. First, let’s figure out the 𝐢th
component of the angular momentum 𝐋.
When we write out the 𝐢-, 𝐣-, and
𝐤-components of the angular momentum, we see, as we look at the first 𝐢th
component, that that component is equal to 𝑟 sub 𝑦 times 𝑣 sub 𝑧 minus 𝑟 sub 𝑧
times 𝑣 sub 𝑦.
But for the position and velocity
vectors given, there is no 𝑧-component, so 𝑣 sub 𝑧 and 𝑟 sub 𝑧 are both
zero. We have both those terms being
zero; the entire 𝐢th component is zero as well.
So now we move on to the 𝐣th
component of our angular momentum. Again, we see 𝑟 sub 𝑧 in the
first term and 𝑣 sub 𝑧 in the second. Since those are both still zero,
the 𝐣th component of 𝐋 is zero too.
Now on to the last component, the
𝐤-direction component: 𝑟 sub 𝑥 minus 𝑣 sub 𝑦 minus 𝑟 sub 𝑦 times 𝑣 sub
𝑥. Looking at our velocity vector, it
has only an 𝑖- or 𝑥-direction component. That means 𝑣 sub 𝑦 is zero.
𝑟 sub 𝑦 is negative 3.0 and 𝑣
sub 𝑥 is 3.0. So our 𝐤-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. 𝐋, the angular momentum of our
system, is equal to the mass 𝑚 multiplied by 9.0 meters squared per second in the
We’re given 𝑚 as 5.0
kilograms. And when we multiply these two
numbers together, we find that the overall angular momentum 𝐋, the product of these
two numbers, is 45 kilograms meters squared per second in the 𝐤-direction. That’s the magnitude and direction
of the system’s angular momentum.
In part two of our problem, we’re
told that a force 𝐅 equals 5.0 newtons in the 𝐣-direction acts on the
particle. And we want to solve for the torque
Let’s 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 𝐫 and
𝐅, and we need to take a cross-product between them to solve for torque 𝛕.
Let’s write out the general
equation for the cross-product between 𝐫 and 𝐅 using our cross-product
formula. With the components of torque
written out, we can now evaluate the 𝐢-, 𝐣-, and 𝐤-components of 𝛕 to see what
their value is.
Looking at the 𝐢th component, we
see that involves 𝐹 sub 𝑧 in the first term and 𝑟 sub 𝑧 in the second term. Looking at our 𝐫 and 𝐅 vectors,
we see that neither of them has a 𝑧-component, so 𝐹 sub 𝑧 and 𝑟 sub 𝑧 are both
zero. That means that the 𝐢-component of
𝛕 is zero as well.
Moving on to the 𝐣-component, we
again see 𝑟 sub 𝑧 in the first term and 𝐹 sub 𝑧 now in the second term. These values are zero and so so is
the 𝐣th component of 𝛕.
Moving to the last component, the
𝐤th component, 𝑟 sub 𝑥 is 2.0, 𝐹 sub 𝑦 is 5.0, 𝑟 sub 𝑦 is negative 3.0, and
𝐹 sub 𝑥 is zero. This means the 𝐤th component of 𝛕
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 𝐤-direction. We’ve now solved for both the
angular momentum and the torque that acts on this particle.