Three particles, each of mass 3.0 kg, are fastened to each other and to a rotation axis by three massless strings, each 0.360 m long, as shown in the Figure. The combination rotates around the rotation axis with an angular velocity of 19.0 rad/s in such a way that the particles remain in a straight line. What is the rotational inertia of the system?

I= 3•mR^2=3•3•(0.36)^2=1.17 kg•m^2,

L=I• ω= 1.17•19=22.23 kg•m^2/s.

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To find the rotational inertia of the system, we need to calculate the moment of inertia for each particle and then add them up.

The moment of inertia for a particle rotating around an axis at a distance r is given by the formula: I = m * r^2, where m is the mass of the particle and r is the distance from the rotation axis.

Given that each particle has a mass of 3.0 kg and the length of the strings connecting each particle to the rotation axis is 0.360 m, the moment of inertia for each particle is:

I = (3.0 kg) * (0.360 m)^2 = 0.3888 kg⋅m^2

Since there are three particles, the total moment of inertia for the system is:

Total I = 3 * 0.3888 kg⋅m^2 = 1.1664 kg⋅m^2

Now, to find the magnitude of the total angular momentum of the three masses, we need to use the formula: L = I * ω, where L is the total angular momentum, I is the rotational inertia of the system, and ω is the angular velocity.

Given that the angular velocity is 19.0 rad/s and the rotational inertia is 1.1664 kg⋅m^2, we can calculate the total angular momentum as follows:

L = (1.1664 kg⋅m^2) * (19.0 rad/s) = 22.168 kg⋅m^2/s

Therefore, the magnitude of the total angular momentum of the three masses is 22.168 kg⋅m^2/s.

To find the rotational inertia of the system, we need to calculate the sum of the individual rotational inertias of the three masses.

The formula for rotational inertia (I) of a point mass rotating about an axis perpendicular to the direction of motion is given by:
I = m * r^2
where m is the mass of the object and r is the distance between the axis of rotation and the point mass.

In this case, each of the three masses has the same mass (m = 3.0 kg) and the same distance from the rotation axis (r = 0.360 m). Therefore, the rotational inertia of each mass is:
I = m * r^2
I = 3.0 kg * (0.360 m)^2
I = 0.3888 kg⋅m^2 (rounded to four decimal places)

Since the three masses are fastened to each other and rotate as a single system, their individual rotational inertias can be added together to find the total rotational inertia of the system:
Total rotational inertia = 3 * Individual rotational inertia
Total rotational inertia = 3 * 0.3888 kg⋅m^2
Total rotational inertia = 1.1664 kg⋅m^2 (rounded to four decimal places)

For the second part of the question, to find the magnitude of the total angular momentum of the three masses, we can use the formula:
L = I * ω
where L represents the angular momentum, I is the rotational inertia, and ω is the angular velocity.

Using the given angular velocity of 19.0 rad/s and the previously calculated total rotational inertia of 1.1664 kg⋅m^2, we can substitute these values into the formula to find the magnitude of the total angular momentum:
L = 1.1664 kg⋅m^2 * 19.0 rad/s
L = 22.1632 kg⋅m^2/s (rounded to four decimal places)

Therefore, the magnitude of the total angular momentum of the three masses is approximately 22.1632 kg⋅m^2/s.