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Grade upto college level Mechanics

Two particles, each with mass m, are fastened to each other and to a rotation axis by two rods, each with length L Land mass M, as shown in Fig. The combination rotates around the rotation axis with angular velocity ω. Obtain an algebraic expression for the rotational inertia of the combination about this axis.
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

Profile image of Shane Macguire
11 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
11 Years ago

To calculate the rotational inertia of this system, we need to consider how each particle contributes to the overall moment of inertia about the rotation axis. In this scenario, we have two particles, each with mass m, connected to a rotation axis by rods of length L and mass M. The rotational inertia, also known as the moment of inertia, is a measure of how difficult it is to change the rotational motion of an object. It depends on the mass distribution relative to the axis of rotation.

Understanding Rotational Inertia

The moment of inertia \( I \) for a system of particles can be defined as:

I = Σ m_i r_i²

Here, \( m_i \) represents the mass of each particle, and \( r_i \) is the perpendicular distance from the axis of rotation to the particle. In our case, we have two particles, each at a distance \( L \) from the rotation axis.

Calculating the Contributions

  • For Particle 1: The distance from the rotation axis is \( L \). Thus, its contribution to the moment of inertia is:
  • I_1 = m L²

  • For Particle 2: Similarly, the distance from the axis of rotation is also \( L \). Therefore, its contribution is the same:
  • I_2 = m L²

Now, we need to account for the rods as well. The mass of each rod is \( M \) and they are also contributing to the overall moment of inertia.

Inertia of the Rods

The moment of inertia of a rod about one end (which is where it is attached to the rotation axis) is given by:

I_{rod} = \frac{1}{3} M L²

Since there are two rods, their combined moment of inertia will be:

I_{rods} = 2 \times \frac{1}{3} M L² = \frac{2}{3} M L²

Combining All Contributions

Now, let’s sum the individual moments of inertia from both particles and both rods:

I_{total} = I_1 + I_2 + I_{rods}

Substituting in our expressions, we have:

I_{total} = m L² + m L² + \frac{2}{3} M L²

Simplifying this gives:

I_{total} = 2m L² + \frac{2}{3} M L²

Final Expression

Thus, the algebraic expression for the rotational inertia of the system about the rotation axis is:

I_{total} = 2m L² + \frac{2}{3} M L²

This final expression accounts for both the masses of the particles and the rods, providing a complete picture of the rotational inertia for the given configuration. Understanding how each component contributes helps in evaluating the dynamics of the system in rotational motion.