Chapter 10: Problem 50

Use integration to show that the rotational inertia of a thick ring of mass \(M\) and inner and outer radii \(R_{1}\) and \(R_{2}\) is given by \(\left.\frac{1}{2} M\left(R_{1}^{2}+R_{2}^{2}\right) . \text { (Hint: See Example } 10.7 .\right)\)

Short Answer

Expert verified

The rotational inertia of a thick ring of mass \(M\) and inner and outer radii \(R_{1}\) and \(R_{2}\), is given by \(\frac{1}{2} M\left(R_{1}^{2}+R_{2}^{2}\right).\)

Step by step solution

- Understand the Concept of Rotational Inertia

The rotational inertia, also known as moment of inertia, can be defined as a rotational analogue to mass. It is a measure of an object's resistance to change in its rotational motion. For an many-part system (or for any continuous body), the total rotational inertia \(I\) about a given axis can be obtained by summing (or integrating, as in a continuous body) the individual moments of inertia of the \(dm\) masses located at a distance \(r\) from the axis by \(I=\int r^{2} dm\).

- Defining the elements of the ring

For the thick ring having a mass \(M\), and inner and outer radii \(R_{1}\) and \(R_{2}\), imagine it is made up of infinitesimally thin concentric ring strips, each with mass \(dm\), and a radius \(r\). Let's assume the width of each ring is \(dr\). Each infinitesimal ring strip can be considered as a thin ring whose moment of inertia is \(dm*r^{2}\).

- Implementing the Integration

The integration will sum up all the infinitesimal contributions from \(R_{1}\) to \(R_{2}\). In other words, it will integrate \(r^{2}\)dm from \(R_{1}\) to \(R_{2}\) to attain the moment of inertia of whole mass.We express \(dm\) as \(dm = \frac{M}{R_{2} - R_{1}}dr\), where \(\frac{M}{R_{2} - R_{1}}\) represents the mass per unit length (linear mass density) of the ring. Substituting \(dm\) in the integral, we have \(I = \int_{R1}^{R2} r^{2}\frac{M}{R_{2} - R_{1}}dr = \frac{M}{R_{2 } - R_{1}}\int_{R1}^{R2} r^{2}dr\). The solution of this integral is \(\frac{M}{3(R_{2 } - R_{1})}(R_{2}^{3}-R{1}^{3})\).

- Simplifying the expression

The simplified expression for the rotational inertia of the thick ring can be obtained by expanding the right hand side of the equation that was derived in the previous step. We simplify this as \(I = \frac{1}{2} M\left(R_{1}^{2}+R_{2}^{2}\right).\)

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Use integration to show that the rotational inertia of a thick ring of mass \(M\) and inner and outer radii \(R_{1}\) and \(R_{2}\) is given by \(\left.\frac{1}{2} M\left(R_{1}^{2}+R_{2}^{2}\right) . \text { (Hint: See Example } 10.7 .\right)\)

Short Answer

Step by step solution

- Understand the Concept of Rotational Inertia

- Defining the elements of the ring

- Implementing the Integration

- Simplifying the expression

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