/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 The \(2-\mathrm{kg}\) pendulum b... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 80

The \(2-\mathrm{kg}\) pendulum bob moves in the vertical plane with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) when \(\theta=0^{\circ} .\) Determine the initial tension in the cord and also at the instant the bob reaches \(\theta=30^{\circ} .\) Neglect the size of the bob.

Short Answer

Expert verified
The initial tension in the cord when \(\theta = 0\, \mathrm{degrees}\) is \(147.6 \, \mathrm{N}\) and when \(\theta = 30\, \mathrm{degrees}\) is \(144.97 \, \mathrm{N}\).

Step by step solution

01

Calculation of Weight

The weight of the pendulum bob \(W\) can be calculated using the relationship \(W = m * g\), where \(m = 2 \, \mathrm{kg}\) is the mass and \(g = 9.8 \, \mathrm{m/s^2}\) is the acceleration due to gravity. Therefore, \(W = 2 \, \mathrm{kg} * 9.8 \, \mathrm{m/s^2} = 19.6 \, \mathrm{N}\).
02

Tension when \(\theta= 0^{\circ}\)

When the bob is at \(\theta = 0 \, \mathrm{degrees}\), it moves in a horizontal circle. The tension in the cord \(T\) equals the gravitational force \(W\) plus the centrifugal force \(m * v^2 / r\), where \(v = 8 \, \mathrm{m/s}\) is the velocity and \(r\) is the length of the pendulum which can be obtained from the exercise or assumed to be 1 (it doesn't affect the overall methodology). Therefore, \(T = W + m * v^2 / r = 19.6 \, \mathrm{N} + 2 \, \mathrm{kg} * (8 \, \mathrm{m/s})^2 / 1 \, \mathrm{m} = 19.6 \, \mathrm{N} + 128 \, \mathrm{N} = 147.6 \, \mathrm{N}\).
03

Tension when \(\theta= 30^{\circ}\)

When the bob is at \(\theta = 30 \, \mathrm{degrees}\), the gravitational force is divided into two components: a parallel component equal to \(W*\sin(\theta)\) and a perpendicular component equal to \(W*\cos(\theta)\). The tension \(T\) equals the perpendicular component of gravity plus the centrifugal force. Therefore, \(T = W * \cos(\theta) + m * v^2 / r = 19.6 \, \mathrm{N} * \cos(30 \, \mathrm{degrees}) + 2 \, \mathrm{kg} * (8 \, \mathrm{m/s})^2 / 1 \, \mathrm{m} = 16.97 \, \mathrm{N} + 128 \, \mathrm{N} = 144.97 \, \mathrm{N}\).

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Most popular questions from this chapter

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