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Chapter 14: Problem 64

A horizontal tree branch is directly above another horizontal tree branch. The elevation of the higher branch is \(9.65 \mathrm{~m}\) above the ground, and the elevation of the lower branch is \(5.99 \mathrm{~m}\) above the ground. Some children decide to use the two branches to hold a tire swing. One end of the tire swing's rope is tied to the higher tree branch so that the bottom of the tire swing is \(0.47 \mathrm{~m}\) above the ground. This swing is thus a restricted pendulum. Start. ing with the complete length of the rope at an initial angle of \(14.2^{\circ}\) with respect to the vertical, how long does it take a child of mass \(29.9 \mathrm{~kg}\) to complete one swing back and forth?

Short Answer

Expert verified
Answer: The time it takes for the child to complete one full swing back and forth is approximately 3.818 seconds.

Step by step solution

01

Calculate the Length of the Rope

The first step is to calculate the length of the rope. We are given that the higher branch is at 9.65 m and the tire swing bottom is 0.47 m above the ground. To do this, subtract the elevations of the bottom of the tire swing from the top branch: Length of the rope = elevation of higher branch - elevation of tire swing above ground L = 9.65 - 0.47 L = \(\boxed{9.18 \mathrm{~m}}\)
02

Calculate the Vertical Distance

Next, we'll calculate the vertical distance between the highest point of the swing and the lowest point. We know that the angle between the rope and the vertical is \(14.2^\circ\), so we can use the sine function to find the vertical distance (h): h = L * sin(angle) h = 9.18 * sin(14.2) h ≈ \(\boxed{2.27 \mathrm{~m}}\)
03

Calculate the Effective Length of the Pendulum

Now, we need to calculate the effective length of the pendulum, which is the vertical distance between the two branches: Effective length, Leff = elevation of higher branch - elevation of lower branch Leff = 9.65 - 5.99 Leff = \(\boxed{3.66 \mathrm{~m}}\)
04

Calculate the Period of the Pendulum

Now that we have the effective length of the pendulum, we can use the formula for the period of a pendulum to calculate the time it takes for one full swing back and forth: T = 2π * √ (Leff / g), where g is the acceleration due to gravity (approximately 9.81 m/s²) T = 2π * √ (3.66/9.81) T ≈ \(\boxed{3.818 \mathrm{~s}}\) The time it takes for the child to complete one swing back and forth is approximately 3.818 seconds.

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Most popular questions from this chapter

The motion of a planet in a circular orbit about a star obeys the equations of simple harmonic motion. If the orbit is observed edge-on, so the planet's motion appears to be onedimensional, the analogy is quite direct: The motion of the planet looks just like the motion of an object on a spring a) Use Kepler's Third Law of planetary motion to determine the "spring constant" for a planet in circular orbit around a star with period \(T\) b) When the planet is at the extremes of its motion observed edge-on, the analogous "spring" is extended to its largest displacement. Using the "spring" analogy, determine the orbital velocity of the planet.

Pendulum A has a bob of mass \(m\) hung from a string of length \(I_{i}\) pendulum \(B\) is identical to \(A\) except its bob has mass \(2 m\). Compare the frequencies of small oscillations of the two pendulums.

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

14.3 A mass that can oscillate without friction on a horizontal surface is attached to a horizontal spring that is pulled to the right \(10.0 \mathrm{~cm}\) and is released from rest. The period of oscillation for the mass is \(5.60 \mathrm{~s}\). What is the speed of the mass at \(t=2.50 \mathrm{~s} ?\) a) \(-2.61 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) b) \(-3.71 \cdot 10^{-2} \mathrm{~m} / \mathrm{s}\) c) \(-3.71 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) d) \(-2.01 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\)

Two springs, each with \(k=125 \mathrm{~N} / \mathrm{m}\), are hung vertically, and \(1.00-\mathrm{kg}\) masses are attached to their ends. One spring is pulled down \(5.00 \mathrm{~cm}\) and released at \(t=0\); the othen is pulled down \(4.00 \mathrm{~cm}\) and released at \(t=0.300 \mathrm{~s}\). Find the phase difference, in degrees, between the oscillations of the two masses and the equations for the vertical displacements of the masses, taking upward to be the positive direction.
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