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Chapter 14: Problem 37

A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass \(M\) and length \(L\) that is pivoted freely about one end. with a solid sphere of the same mass, \(M,\) and a radius of \(L / 2\) centered about the free end of the rod. a) Obtain an expression for the moment of inertia of the pendulum about its pivot point as a function of \(M\) and \(L\). b) Obtain an expression for the period of the pendulum for small oscillations.

Short Answer

Expert verified
Answer: The moment of inertia of the pendulum is \(I_{total} = \frac{8}{15}ML^2\). The period of the pendulum for small oscillations is \(T = 2\pi\sqrt{\frac{8L}{15g}}\).

Step by step solution

01

Find the moment of inertia of the rod

The moment of inertia for a uniform thin rod about one end is given by the equation \(I_{rod} = \frac{1}{3}ML^2\), where \(M\) is the mass of the rod and \(L\) is the length of the rod.
02

Find the moment of inertia of the sphere

The moment of inertia for a solid sphere about a diameter is \(I_{sphere} = \frac{2}{5}MR^2\), where \(M\) is the mass of the sphere and \(R\) is the radius of the sphere. For this physical pendulum, the radius of the sphere is \(L / 2\). Therefore, \(I_{sphere} = \frac{2}{5}M(\frac{L}{2})^2\).
03

Calculate the parallel-axis theorem for the sphere

As the sphere is pivoted about the endpoint of the rod, we will use the parallel-axis theorem to find the total moment of inertia about the pivot. The parallel-axis theorem states that \(I = I_{cm} + Md^2\), where \(I_{cm}\) is the moment of inertia of the sphere about its center of mass, \(M\) is the mass of the sphere, and \(d\) is the perpendicular distance between the axes (\(L\) in this case because the sphere is centered about the free end of the rod). So, \(I_{sphere} = \frac{2}{5}M(\frac{L}{2})^2 + M L^2\).
04

Calculate the total moment of inertia of the pendulum

Now, we can find the total moment of inertia for the pendulum by adding the moment of inertia of the rod and the sphere together. \(I_{total} = I_{rod} + I_{sphere} = \frac{1}{3}ML^2 + \frac{2}{5}M(\frac{L}{2})^2 + M L^2\).
05

Simplify the equation for the total moment of inertia

The total moment of inertia equation should be simplified. \(I_{total} = \frac{1}{3}ML^2 + \frac{1}{5}ML^2 + ML^2 = (\frac{1}{3} + \frac{1}{5} + 1)ML^2 = \frac{8}{15}ML^2\).
06

Find the period for small oscillations

For small oscillations of a physical pendulum, the period \(T\) can be found using the equation \(T = 2\pi\sqrt{\frac{I_{total}}{MgR}}\). In this case, \(I_{total}\) is the moment of inertia we found in the previous step, \(M\) is the mass of the pendulum (both rod and sphere), \(g\) is the gravitational acceleration, and \(R\) is the perpendicular distance from the pivot point to the center of mass of the pendulum. As the sphere is at the end of the rod and the mass of the rod is evenly distributed along it, the center of mass of the pendulum is at the midpoint of the rod or \(\frac{L}{2}\) away from the pivot. Hence, \(T = 2\pi\sqrt{\frac{\frac{8}{15}ML^2}{MgL/2}}\).
07

Simplify the period equation

Finally, we can simplify the equation for the period. \(T = 2\pi\sqrt{\frac{\frac{8}{15}ML^2}{\frac{1}{2}MgL}} = 2\pi\sqrt{\frac{8L}{15g}}\). The moment of inertia of the pendulum is \(I_{total} = \frac{8}{15}ML^2\). The period of the pendulum for small oscillations is \(T = 2\pi\sqrt{\frac{8L}{15g}}\).

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