Two skaters, each of mass 70 kg, skate at speeds of 4 m/s in opposite directions along

parallel lines 1.5 m apart. As they are about to pass one another they join hands and go
into circular paths about their common center of mass.
A third skater is skating at 2 m/s along a line parallel to the initial directions of the
other two and 6 m off to the side of the track of the nearer one. From his standpoint,
what is the total angular momentum of the other two skaters as they rotate?

To solve this problem, we need to find the total angular momentum of the two skaters as they rotate. The angular momentum of an object is defined as the product of its moment of inertia and its angular velocity.

Let's break down the problem step by step:

Step 1: Find the moment of inertia of the two skaters about their common center of mass.
The moment of inertia depends on the distribution of mass and the rotation axis. In this case, since the skaters join hands and rotate about their common center of mass, we can approximate them as a single object with a moment of inertia given by:

I = (m1 * r1^2) + (m2 * r2^2)

Where:
m1 = mass of the first skater
r1 = distance of the first skater from the center of mass
m2 = mass of the second skater
r2 = distance of the second skater from the center of mass

In our case, both skaters have the same mass (m1 = m2 = 70 kg) and the distance between them is given as 1.5 m (r1 = r2 = 1.5 m).

I = (70 kg * 1.5 m)^2 + (70 kg * 1.5 m)^2 = 2 * (70 kg * 1.5 m)^2

Step 2: Find the angular velocity of the rotation.
The angular velocity can be calculated by dividing the tangential velocity by the radius of the circular path. In this case, the tangential velocity is the sum of each skater's speed (since they are moving in opposite directions):

v = v1 + v2

Where:
v1 = speed of the first skater = 4 m/s
v2 = speed of the second skater = 4 m/s

v = 4 m/s + (-4 m/s) = 0 m/s

The radius of the circular path is the distance between the skaters' center of mass and the third skater:

r = 6 m

ω = v / r = 0 m/s / 6 m = 0 rad/s

Step 3: Calculate the total angular momentum.
The total angular momentum of the two skaters is given by the product of the moment of inertia and the angular velocity:

L = I * ω

L = (2 * (70 kg * 1.5 m)^2) * 0 rad/s
L = 0

Therefore, from the standpoint of the third skater, the total angular momentum of the other two skaters as they rotate is zero.

To find the total angular momentum of the two skaters as they rotate from the third skater's standpoint, we need to calculate the individual angular momenta of each skater and then add them together. The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

First, let's consider the two skaters who joined hands and are rotating about their common center of mass. Since they joined hands, we can treat them as a single system with a combined mass of 2 * 70 kg = 140 kg.

To calculate their moment of inertia, we need to know the radius of their circular path. From the problem statement, we know that they are 1.5 m apart when they join hands. Therefore, their circular path has a radius of 0.75 m (half the distance between them).

The moment of inertia for a system rotating about an axis passing through its center of mass is given by the formula I = MR^2, where M is the total mass of the system and R is the radius of rotation. Therefore, the moment of inertia for the two skaters is I = 140 kg * (0.75 m)^2 = 78.75 kg*m^2.

Next, we need to find the angular velocity of the rotating skaters. Since we know their speeds when they join hands, we can use the relationship v = ωR, where v is the linear velocity, ω is the angular velocity, and R is the radius of rotation. In this case, the skaters were moving at 4 m/s, and the radius of rotation is 0.75 m. So, ω = v / R = 4 m/s / 0.75 m = 5.33 rad/s.

Now, we can calculate the angular momentum for the rotating skaters using the formula L = Iω. Therefore, L = 78.75 kg*m^2 * 5.33 rad/s = 419.33 kg*m^2/s.

Finally, we need to consider the third skater who is skating along a line parallel to the initial directions of the other two and 6 m off to the side of the track of the nearer one. Since this skater is not participating in the circular motion, their angular momentum is zero.

So, from the third skater's standpoint, the total angular momentum of the other two skaters as they rotate is 419.33 kg*m^2/s.

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