How Does The Force Of Gravity Between Two Bodies Change When The Distance Between Them Doubles?

Learning Objectives

Past the end of this section, you will be able to:

• State the forces that act on a simple pendulum
• Determine the athwart frequency, frequency, and period of a unproblematic pendulum in terms of the length of the pendulum and the acceleration due to gravity
• Define the period for a physical pendulum
• Define the flow for a torsional pendulum

Pendulums are in common usage. Grandfather clocks apply a pendulum to keep fourth dimension and a pendulum can be used to mensurate the dispatch due to gravity. For small displacements, a pendulum is a elementary harmonic oscillator.

The Simple Pendulum

A unproblematic pendulum is defined to have a point mass, also known as the pendulum bob , which is suspended from a cord of length L with negligible mass ((Effigy)). Here, the just forces acting on the bob are the force of gravity (i.east., the weight of the bob) and tension from the string. The mass of the cord is assumed to be negligible every bit compared to the mass of the bob.

Figure 15.20 A uncomplicated pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which event in a net force of [latex] \text{−}mg\text{sin}\,\theta [/latex] toward the equilibrium position—that is, a restoring force.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the cord L times the component of the cyberspace strength that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the contrary direction of the angular displacement:

[latex] \begin{array}{ccc}\hfill \tau & =\hfill & \text{−}L(mg\,\text{sin}\,\theta );\hfill \\ \hfill I\alpha & =\hfill & \text{−}L(mg\,\text{sin}\,\theta );\hfill \\ \hfill I\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}L(mg\,\text{sin}\,\theta );\hfill \\ \hfill m{L}^{2}\frac{{d}^{2}\theta }{d{t}^{2}}& =\hfill & \text{−}L(mg\,\text{sin}\,\theta );\hfill \\ \hfill \frac{{d}^{2}\theta }{d{t}^{two}}& =\hfill & -\frac{g}{50}\text{sin}\,\theta .\hfill \stop{array} [/latex]

The solution to this differential equation involves advanced calculus, and is across the scope of this text. Simply note that for small angles (less than 15 degrees), [latex] \text{sin}\,\theta [/latex] and [latex] \theta [/latex] differ by less than 1%, and so we tin employ the modest angle approximation [latex] \text{sin}\,\theta \approx \theta . [/latex] The angle [latex] \theta [/latex] describes the position of the pendulum. Using the small angle approximation gives an approximate solution for modest angles,

[latex] \frac{{d}^{two}\theta }{d{t}^{2}}=-\frac{one thousand}{L}\theta . [/latex]

Because this equation has the aforementioned form as the equation for SHM, the solution is like shooting fish in a barrel to detect. The angular frequency is

[latex] \omega =\sqrt{\frac{g}{L}} [/latex]

and the period is

[latex] T=2\pi \sqrt{\frac{L}{g}}. [/latex]

The period of a simple pendulum depends on its length and the acceleration due to gravity. The catamenia is completely independent of other factors, such as mass and the maximum deportation. As with elementary harmonic oscillators, the flow T for a pendulum is well-nigh independent of amplitude, particularly if [latex] \theta [/latex] is less than about [latex] 15\text{°}. [/latex] Even simple pendulum clocks tin exist finely adjusted and remain accurate.

Note the dependence of T on one thousand. If the length of a pendulum is precisely known, it can actually exist used to measure the dispatch due to gravity, as in the following example.

Example

Measuring Acceleration due to Gravity by the Flow of a Pendulum

What is the acceleration due to gravity in a region where a unproblematic pendulum having a length 75.000 cm has a period of i.7357 s?

Strategy

We are asked to observe g given the flow T and the length L of a pendulum. We can solve [latex] T=2\pi \sqrt{\frac{L}{chiliad}} [/latex] for g, bold just that the bending of deflection is less than [latex] 15\text{°} [/latex].

Solution

1. Square [latex] T=ii\pi \sqrt{\frac{Fifty}{yard}} [/latex] and solve for m:

   [latex] g=4{\pi }^{2}\frac{L}{{T}^{2}}. [/latex]

2. Substitute known values into the new equation:

   [latex] g=4{\pi }^{two}\frac{0.75000\,\text{1000}}{{(i.7357\,\text{southward})}^{two}}. [/latex]

3. Calculate to detect g:

   [latex] g=nine.8281{\,\text{m/s}}^{two}. [/latex]

Significance

This method for determining g can be very accurate, which is why length and period are given to five digits in this case. For the precision of the approximation [latex] \text{sin}\,\theta \approx \theta [/latex] to be better than the precision of the pendulum length and menses, the maximum deportation angle should be kept below nearly [latex] 0.v\text{°} [/latex].

Check Your Understanding

An engineer builds two unproblematic pendulums. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum two has a bob with a mass of 100 kg. Draw how the motion of the pendulums will differ if the bobs are both displaced by [latex] 12\text{°} [/latex].

Testify Solution

The movement of the pendulums will not differ at all because the mass of the bob has no result on the motion of a simple pendulum. The pendulums are only affected by the flow (which is related to the pendulum's length) and past the acceleration due to gravity.

Physical Pendulum

Whatever object can oscillate like a pendulum. Consider a coffee mug hanging on a claw in the pantry. If the mug gets knocked, it oscillates dorsum and forth like a pendulum until the oscillations die out. We accept described a simple pendulum as a point mass and a string. A physical pendulum is whatsoever object whose oscillations are like to those of the elementary pendulum, but cannot exist modeled as a bespeak mass on a string, and the mass distribution must be included into the equation of motion.

Equally for the simple pendulum, the restoring strength of the physical pendulum is the forcefulness of gravity. With the unproblematic pendulum, the forcefulness of gravity acts on the center of the pendulum bob. In the case of the concrete pendulum, the force of gravity acts on the center of mass (CM) of an object. The object oscillates near a point O. Consider an object of a generic shape as shown in (Figure).

Effigy xv.21 A physical pendulum is whatsoever object that oscillates every bit a pendulum, just cannot be modeled equally a indicate mass on a string. The forcefulness of gravity acts on the center of mass (CM) and provides the restoring force that causes the object to oscillate. The minus sign on the component of the weight that provides the restoring force is present because the force acts in the opposite direction of the increasing angle [latex] \theta [/latex].

When a concrete pendulum is hanging from a signal but is complimentary to rotate, information technology rotates considering of the torque practical at the CM, produced past the component of the object's weight that acts tangent to the motion of the CM. Taking the counterclockwise direction to be positive, the component of the gravitational forcefulness that acts tangent to the movement is [latex] \text{−}mg\,\text{sin}\,\theta [/latex]. The minus sign is the result of the restoring force acting in the opposite direction of the increasing bending. Recall that the torque is equal to [latex] \overset{\to }{\tau }=\overset{\to }{r}\,×\,\overset{\to }{F} [/latex]. The magnitude of the torque is equal to the length of the radius arm times the tangential component of the strength practical, [latex] |\tau |=rF\text{sin}\,\theta [/latex]. Here, the length L of the radius arm is the distance betwixt the point of rotation and the CM. To clarify the movement, get-go with the net torque. Like the simple pendulum, consider only small angles so that [latex] \text{sin}\,\theta \approx \theta [/latex]. Recall from Fixed-Axis Rotation on rotation that the internet torque is equal to the moment of inertia [latex] I=\int {r}^{ii}dm [/latex] times the athwart dispatch [latex] \alpha , [/latex] where [latex] \blastoff =\frac{{d}^{2}\theta }{d{t}^{2}} [/latex]:

[latex] I\alpha ={\tau }_{\text{cyberspace}}=L(\text{−}mg)\text{sin}\,\theta . [/latex]

Using the small bending approximation and rearranging:

[latex] \begin{assortment}{ccc}\hfill I\alpha & =\hfill & \text{−}50(mg)\theta ;\hfill \\ \hfill I\frac{{d}^{two}\theta }{d{t}^{ii}}& =\hfill & \text{−}L(mg)\theta ;\hfill \\ \hfill \frac{{d}^{two}\theta }{d{t}^{2}}& =\hfill & \text{−}(\frac{mgL}{I})\theta .\hfill \end{array} [/latex]

Once once more, the equation says that the second time derivative of the position (in this instance, the angle) equals minus a constant [latex] (-\frac{mgL}{I}) [/latex] times the position. The solution is

[latex] \theta (t)=\text{Θ}\text{cos}(\omega t+\varphi ), [/latex]

where [latex] \text{Θ} [/latex] is the maximum angular displacement. The athwart frequency is

[latex] \omega =\sqrt{\frac{mgL}{I}}. [/latex]

The catamenia is therefore

[latex] T=2\pi \sqrt{\frac{I}{mgL}}. [/latex]

Annotation that for a simple pendulum, the moment of inertia is [latex] I=\int {r}^{2}dm=grand{L}^{two} [/latex] and the menstruum reduces to [latex] T=2\pi \sqrt{\frac{Fifty}{thousand}} [/latex].

Example

Reducing the Swaying of a Skyscraper

In extreme conditions, skyscrapers tin sway up to ii meters with a frequency of upward to 20.00 Hz due to high winds or seismic activeness. Several companies take developed physical pendulums that are placed on the top of the skyscrapers. As the skyscraper sways to the right, the pendulum swings to the left, reducing the sway. Assuming the oscillations have a frequency of 0.50 Hz, design a pendulum that consists of a long axle, of constant density, with a mass of 100 metric tons and a pivot point at one finish of the beam. What should be the length of the beam?

Strategy

We are asked to find the length of the physical pendulum with a known mass. Nosotros first need to find the moment of inertia of the beam. Nosotros tin so use the equation for the period of a concrete pendulum to find the length.

Solution

1. Observe the moment of inertia for the CM:
2. Use the parallel axis theorem to detect the moment of inertia about the point of rotation:

   [latex] I={I}_{\text{CM}}+{\frac{L}{4}}^{2}M=\frac{i}{12}Chiliad{L}^{2}+\frac{1}{4}M{50}^{2}=\frac{1}{3}M{L}^{2}. [/latex]

3. The menses of a concrete pendulum has a catamenia of [latex] T=2\pi \sqrt{\frac{I}{mgL}} [/latex]. Use the moment of inertia to solve for the length L:

   [latex] \brainstorm{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{I}{MgL}}=2\pi \sqrt{\frac{\frac{1}{three}M{L}^{2}}{MgL}}=ii\pi \sqrt{\frac{L}{3g}};\hfill \\ \hfill L& =\hfill & 3g{(\frac{T}{2\pi })}^{ii}=3(9.8\frac{\text{chiliad}}{{\text{s}}^{2}}){(\frac{2\,\text{s}}{2\pi })}^{2}=2.98\,\text{m}\text{.}\hfill \cease{assortment} [/latex]

Significance

There are many ways to reduce the oscillations, including modifying the shape of the skyscrapers, using multiple concrete pendulums, and using tuned-mass dampers.

Torsional Pendulum

A torsional pendulum consists of a rigid body suspended by a light wire or spring ((Figure)). When the body is twisted some pocket-sized maximum angle [latex] (\text{Θ}) [/latex] and released from rest, the body oscillates between [latex] (\theta =+\text{Θ}) [/latex] and [latex] (\theta =\text{−}\,\text{Θ}) [/latex]. The restoring torque is supplied by the shearing of the string or wire.

Figure 15.22 A torsional pendulum consists of a rigid trunk suspended by a cord or wire. The rigid body oscillates between [latex] \theta =+\text{Θ} [/latex] and [latex] \theta =\text{−}\text{Θ} [/latex].

The restoring torque tin exist modeled as being proportional to the angle:

[latex] \tau =\text{−}\kappa \theta . [/latex]

The variable kappa [latex] (\kappa ) [/latex] is known as the torsion constant of the wire or string. The minus sign shows that the restoring torque acts in the opposite management to increasing angular displacement. The net torque is equal to the moment of inertia times the athwart acceleration:

[latex] \begin{array}{}\\ I\frac{{d}^{2}\theta }{d{t}^{two}}=\text{−}\kappa \theta ;\hfill \\ \frac{{d}^{2}\theta }{d{t}^{2}}=-\frac{\kappa }{I}\theta .\hfill \end{array} [/latex]

This equation says that the second time derivative of the position (in this case, the angle) equals a negative constant times the position. This looks very similar to the equation of movement for the SHM [latex] \frac{{d}^{2}x}{d{t}^{2}}=-\frac{k}{thousand}x [/latex], where the period was found to be [latex] T=ii\pi \sqrt{\frac{1000}{k}} [/latex]. Therefore, the menses of the torsional pendulum can be constitute using

[latex] T=two\pi \sqrt{\frac{I}{\kappa }}. [/latex]

The units for the torsion constant are [latex] [\kappa ]=\text{Northward-m}=(\text{kg}\frac{\text{grand}}{{\text{southward}}^{2}})\text{m}=\text{kg}\,\frac{{\text{k}}^{2}}{{\text{southward}}^{ii}} [/latex] and the units for the moment of inertial are [latex] [I]={\text{kg-m}}^{two}, [/latex] which evidence that the unit for the menses is the 2d.

Case

Measuring the Torsion Abiding of a String

A rod has a length of [latex] 50=0.thirty\,\text{k} [/latex] and a mass of 4.00 kg. A string is attached to the CM of the rod and the organisation is hung from the ceiling ((Figure)). The rod is displaced 10 degrees from the equilibrium position and released from balance. The rod oscillates with a flow of 0.5 s. What is the torsion constant [latex] \kappa [/latex]?

Figure xv.23 (a) A rod suspended by a string from the ceiling. (b) Finding the rod's moment of inertia.

Strategy

We are asked to notice the torsion constant of the string. We outset demand to find the moment of inertia.

Solution

1. Find the moment of inertia for the CM:

   [latex] {I}_{\text{CM}}=\int {ten}^{2}dm={\int }_{\text{−}Fifty\text{/}two}^{+L\text{/}two}{x}^{2}\lambda dx=\lambda {[\frac{{x}^{iii}}{3}]}_{\text{−}Fifty\text{/}2}^{+Fifty\text{/}ii}=\lambda \frac{two{50}^{3}}{24}=(\frac{M}{L})\frac{2{L}^{3}}{24}=\frac{1}{12}M{L}^{two}. [/latex]

2. Calculate the torsion constant using the equation for the period:

   [latex] \begin{array}{ccc}\hfill T& =\hfill & ii\pi \sqrt{\frac{I}{\kappa }};\hfill \\ \hfill \kappa & =\hfill & I{(\frac{2\pi }{T})}^{two}=(\frac{1}{12}G{L}^{two}){(\frac{2\pi }{T})}^{2};\hfill \\ & =\hfill & (\frac{1}{12}(4.00\,\text{kg}){(0.thirty\,\text{one thousand})}^{2}){(\frac{two\pi }{0.50\,\text{s}})}^{2}=4.73\,\text{N}·\text{k}\text{.}\hfill \terminate{array} [/latex]

Significance

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

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 [latex] 15\text{°} [/latex]. The menses of a simple pendulum is [latex] T=ii\pi \sqrt{\frac{L}{g}} [/latex], where L is the length of the cord and g is the acceleration due to gravity.
• The period of a physical pendulum [latex] T=2\pi \sqrt{\frac{I}{mgL}} [/latex] tin can be constitute if the moment of inertia is known. The length between the point of rotation and the center of mass is L.
• The catamenia of a torsional pendulum [latex] T=two\pi \sqrt{\frac{I}{\kappa }} [/latex] can be found if the moment of inertia and torsion abiding are known.

Conceptual Questions

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

A pendulum clock works by measuring the catamenia 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 as well fast or as well slow in the summer? What about the wintertime?

Show Solution

The flow of the pendulum is [latex] T=2\pi \sqrt{Fifty\text{/}g}. [/latex] In summer, the length increases, and the menses increases. If the menses should exist i second, but period is longer than 1 second in the summertime, 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 stage shift, the position of an object may exist modeled as a cosine or sine function. If given the choice, which function would you lot choose? Bold that the phase shift is zero, what are the initial conditions of function; that is, the initial position, velocity, and dispatch, when using a sine function? How most when a cosine office is used?

Problems

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

Some people think a pendulum with a period of i.00 s tin can be driven with "mental energy" or psycho kinetically, because its flow is the same as an average heartbeat. True or not, what is the length of such a pendulum?

What is the menses of a 1.00-chiliad-long pendulum?

How long does information technology take a kid on a swing to complete ane swing if her center of gravity is four.00 m below the pivot?

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

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

(a) A pendulum that has a period of iii.00000 s and that is located where the acceleration due to gravity is [latex] 9.79\,{\text{m/s}}^{2} [/latex] is moved to a location where the dispatch due to gravity is [latex] nine.82\,{\text{m/s}}^{two} [/latex]. What is its new menstruum? (b) Explicate why and so many digits are needed in the value for the period, based on the relation betwixt the flow and the acceleration due to gravity.

A pendulum with a menstruation of 2.00000 s in one location ([latex] g=9.80{\text{m/s}}^{ii} [/latex]) is moved to a new location where the period is at present 1.99796 due south. What is the acceleration due to gravity at its new location?

Show Solution

[latex] 9.82002\,{\text{grand/s}}^{2} [/latex]

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

Glossary

concrete pendulum

any extended object that swings like a pendulum

simple pendulum

point mass, chosen a pendulum bob, fastened to a well-nigh massless string

torsional pendulum

any suspended object that oscillates by twisting its suspension

Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/15-4-pendulums/

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