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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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Most popular questions from this chapter

A grandfather clock uses a pendulum and a weight. The pendulum has a period of \(2.00 \mathrm{~s}\), and the mass of the bob is 250. \(\mathrm{g}\). The weight slowly falls, providing the energy to overcome the damping of the pendulum due to friction. The weight has a mass of \(1.00 \mathrm{~kg}\), and it moves down \(25.0 \mathrm{~cm}\) every day. Find \(Q\) for this clock. Assume that the amplitude of the oscillation of the pendulum is \(10.0^{\circ}\)

Consider two identical oscillators, each with spring constant \(k\) and mass \(m\), in simple harmonic motion. One oscillator is started with initial conditions \(x_{0}\) and \(v_{j}\) the other starts with slightly different conditions, \(x_{0}+\delta x\) and \(v_{0}+\delta v_{1}\) a) Find the difference in the oscillators' positions, \(x_{1}(t)-x_{2}(t)\) for all t. b) This difference is bounded; that is, there exists a constant \(C\) independent of time, for which \(\left|x_{1}(t)-x_{2}(t)\right| \leq C\) holds for all \(t\). Find an expression for \(C\). What is the best bound, that is, the smallest value of \(C\) that works? (Note: An important characteristic of chaotic systems is exponential sensitivity to initial conditions; the difference in position of two such systems with slightly different initial conditions grows exponentially with time. You have just shown that an oscillator in simple harmonic motion is not a chaotic system.)

Mass-spring systems and pendulum systems can both be used in mechanical timing devices. What are the advantages of using one type of system rather than the othes in a device designed to generate reproducible time measurements over an extended period of time?

You have a linear (following Hooke's Law) spring with an unknown spring constant, a standard mass, and a timer. Explain carefully how you could most practically use these to measure masses in the absence of gravity. Be as quantitative as you can. Regard the mass of the spring as negligible

A Foucault pendulum displayed in a museum is typically quite long, making the effect easier to see. Consider a Foucault pendulum of length \(15 \mathrm{~m}\) with a 110 -kg brass bob. It is set to swing with an amplitude of \(3.5^{\circ}\) a) What is the period of the pendulum? b) What is the maximum kinetic energy of the pendulum? c) What is the maximum speed of the pendulum?
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