# 15.4 Pendulums  (Page 4/7)

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## Measuring the torsion constant of a string

A rod has a length of $l=0.30\phantom{\rule{0.2em}{0ex}}\text{m}$ and a mass of 4.00 kg. A string is attached to the CM of the rod and the system is hung from the ceiling ( [link] ). The rod is displaced 10 degrees from the equilibrium position and released from rest. The rod oscillates with a period of 0.5 s. What is the torsion constant $\kappa$ ?

## Strategy

We are asked to find the torsion constant of the string. We first need to find the moment of inertia.

## Solution

1. Find the moment of inertia for the CM:
${I}_{\text{CM}}=\int {x}^{2}dm={\int }_{\text{−}L\text{/}2}^{+L\text{/}2}{x}^{2}\lambda dx=\lambda {\left[\frac{{x}^{3}}{3}\right]}_{\text{−}L\text{/}2}^{+L\text{/}2}=\lambda \frac{2{L}^{3}}{24}=\left(\frac{M}{L}\right)\frac{2{L}^{3}}{24}=\frac{1}{12}M{L}^{2}.$
2. Calculate the torsion constant using the equation for the period:
$\begin{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{\kappa }};\hfill \\ \hfill \kappa & =\hfill & I{\left(\frac{2\pi }{T}\right)}^{2}=\left(\frac{1}{12}M{L}^{2}\right){\left(\frac{2\pi }{T}\right)}^{2};\hfill \\ & =\hfill & \left(\frac{1}{12}\left(4.00\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(0.30\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}\right){\left(\frac{2\pi }{0.50\phantom{\rule{0.2em}{0ex}}\text{s}}\right)}^{2}=4.73\phantom{\rule{0.2em}{0ex}}\text{N}·\text{m}\text{.}\hfill \end{array}$

## Significance

Like the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the period.

## Summary

• A mass m suspended by a wire of length L and negligible mass is a simple pendulum and undergoes SHM for amplitudes less than about $15\text{°}$ . The period of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$ , where L is the length of the string and g is the acceleration due to gravity.
• The period of a physical pendulum $T=2\pi \sqrt{\frac{I}{mgL}}$ can be found if the moment of inertia is known. The length between the point of rotation and the center of mass is L .
• The period of a torsional pendulum $T=2\pi \sqrt{\frac{I}{\kappa }}$ can be found if the moment of inertia and torsion constant are known.

## Conceptual questions

Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.

A pendulum clock works by measuring the period of a pendulum. In the springtime the clock runs with perfect time, but in the summer and winter the length of the pendulum changes. When most materials are heated, they expand. Does the clock run too fast or too slow in the summer? What about the winter?

The period of the pendulum is $T=2\pi \sqrt{L\text{/}g}.$ In summer, the length increases, and the period increases. If the period should be one second, but period is longer than one second in the summer, it will oscillate fewer than 60 times a minute and clock will run slow. In the winter it will run fast.

With the use of a phase shift, the position of an object may be modeled as a cosine or sine function. If given the option, which function would you choose? Assuming that the phase shift is zero, what are the initial conditions of function; that is, the initial position, velocity, and acceleration, when using a sine function? How about when a cosine function is used?

## Problems

What is the length of a pendulum that has a period of 0.500 s?

Some people think a pendulum with a period of 1.00 s can be driven with “mental energy” or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?

24.8 cm

What is the period of a 1.00-m-long pendulum?

How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

4.01 s

The pendulum on a cuckoo clock is 5.00-cm long. What is its frequency?

Two parakeets sit on a swing with their combined CMs 10.0 cm below the pivot. At what frequency do they swing?

1.58 s

(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is $9.79\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ is moved to a location where the acceleration due to gravity is $9.82\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ . What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.

A pendulum with a period of 2.00000 s in one location ( $g=9.80{\text{m/s}}^{2}$ ) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?

$9.82002\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$

(a) What is the effect on the period of a pendulum if you double its length? (b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?

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