/*! 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 84 A 200-g steel ball is tied to a ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 84

A 200-g steel ball is tied to a 2.00-m "massless" string and hung from the ceiling to make a pendulum, and then, the ball is brought to a position making a \(30^{\circ}\) angle with the vertical direction and released from rest. Ignoring the effects of the air resistance, find the speed of the ball when the string (a) is vertically down, (b) makes an angle of \(20^{\circ}\) with the vertical and (c) makes an angle of \(10^{\circ}\) with the vertical.

Short Answer

Expert verified
The speed of the ball when the string is vertically down, makes an angle of \(20^{\circ}\) with the vertical and makes an angle of \(10^{\circ}\) with the vertical are calculated by equating the potential energy (mgh) and the kinetic energy \(\frac{1}{2}mv^2\) at the initial and respective points. The exact values can be found by substituting the given values into the energy conservation equations derived in Steps 2, 3 and 4 above.

Step by step solution

01

Analyzing and Understanding the Problem

The ball is released rest and initially it's at a height greater than its final height. So, it starts with certain potential energy and ends up with kinetic energy. The total mechanical energy of the system remains constant, as there's no energy being added or subtracted. Therefore, the potential energy at the top must equal the sum of the potential energy and kinetic energy at any other point.
02

Find the speed of the ball when the string is vertically down

When the string is vertically down, the maximum conversion of potential energy to kinetic energy has occurred. Hence the potential energy at 30 degrees is all converted to kinetic energy. We use the equation of conservation energy at its lowest position: \[mgh_{30} = \frac{1}{2}mv_{bottom}^2\]where \(h_{30} = L-L\cos{30}\), \(m = 0.2 kg\), \(L = 2.0 m\) and \(g = 9.81 m/s^2\). Solve the equation for \(v_{bottom}\) to get our answer.
03

Find the speed of the ball when the string makes an angle of \(20^{\circ}\) with the vertical

In this case, some of the original potential energy is still potential energy and the rest is converted to kinetic energy. We use the equation of conservation of energy at \(20^{\circ}\):\[mgh_{30} = mgh_{20} + \frac{1}{2}mv_{20}^2\]where \(h_{20} = L - L\cos{20}\). We solve the equation for \(v_{20}\) to get the speed.
04

Find the speed of the ball when the string makes an angle of \(10^{\circ}\) with the vertical.

Similarly, we use the equation of conservation of energy at \(10^{\circ}\): \[mgh_{30} = mgh_{10} + \frac{1}{2}mv_{10}^2\]where \(h_{10} = L - L\cos{10}\). We solve the equation for \(v_{10}\) to get the speed.
05

Conclusion

The speed of the ball is calculated by equating the potential energy and kinetic energy at the highest point and the respective points. By using the conservation of energy principle, we get the speed of the ball when the string is vertically down, makes an angle of \(20^{\circ}\) with the vertical and makes an angle of \(10^{\circ}\) with the vertical.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Motion
A pendulum consists of a weight suspended from a fixed point that swings back and forth. In our exercise, the pendulum is a 200-gram steel ball tied to a string, hanging from the ceiling. The motion of a pendulum is a classic example of simple harmonic motion. This motion can be described using physics principles, without considering friction or air resistance in this theoretical scenario.

When the ball is released from its initial angle with the vertical, it will descend under gravity, swinging down and back up. Each swing is a result of the conversion between potential and kinetic energy, which is why understanding these energy forms is essential in analyzing pendulum motion.
  • Two key positions for a pendulum are its highest point and lowest point.
  • At the highest point, the ball has the maximum potential energy and zero kinetic energy as it is momentarily at rest.
  • At the lowest point, potential energy is at a minimum, and the ball reaches its maximum speed with maximum kinetic energy.
Understanding the conservation of energy is crucial to predicting the pendulum's speed at any given point.
Kinetic Energy
Kinetic energy is the energy of motion. For the pendulum, kinetic energy is at its highest when the ball is moving fastest. In our problem, this occurs when the string is vertically down, because that's when almost all of the potential energy has been converted into kinetic energy.

Using the equation for kinetic energy, we can express it as:\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object (in this case, the ball), and \( v \) is its velocity. To find the velocity of the ball, we equate its initial potential energy to its kinetic energy at specific points, such as when the pendulum is vertical or at certain angles.
  • The speed is calculated using the conservation of energy principle.
  • Kinetic energy increases as the pendulum moves downward from its highest point.
Understanding kinetic energy is a helpful way to determine the speed of the pendulum's swings.
Potential Energy
Potential energy is stored energy due to an object's position or height. In a pendulum, this energy is highest at the start of the motion, when the ball is held at a certain angle with the vertical direction. At this point, all energy is potential because the ball is momentarily at rest.

The potential energy when the ball is at the initial angle can be calculated with the equation:\[ PE = mgh \]Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height. The height \( h \) can be determined by the vertical component of the pendulum's length at different angles.
  • As the pendulum swings down, potential energy decreases while kinetic energy increases.
  • At the lowest point, potential energy is at its lowest because the height difference from the initial position is maximized.
By understanding potential energy, we can explain why energy conservation helps determine pendulum motion energetics.

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Most popular questions from this chapter

(a) How high a hill can a car coast up (engines disengaged) if work done by friction is negligible and its initial speed is \(110 \mathrm{km} / \mathrm{h}\) ? (b) If, in actuality, a \(750-\mathrm{kg}\) car with an initial speed of \(110 \mathrm{km} / \mathrm{h}\) is observed to coast up a hill to a height \(22.0 \mathrm{m}\) above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope of \(2.5^{\circ}\) above the horizontal?

A child (32 kg) jumps up and down on a trampoline. The trampoline exerts a spring restoring force on the child with a constant of \(5000 \mathrm{N} / \mathrm{m}\). At the highest point of the bounce, the child is \(1.0 \mathrm{m}\) above the level surface of the trampoline. What is the compression distance of the trampoline? Neglect the bending of the legs or any transfer of energy of the child into the trampoline while jumping.

A particle of mass 4.0 kg is constrained to move along the \(x\) -axis under a single force \(F(x)=-c x^{3},\) where \(c=8.0 \mathrm{N} / \mathrm{m}^{3} .\) The particle's speed at \(A,\) where \(x_{A}=1.0 \mathrm{m},\) is \(6.0 \mathrm{m} / \mathrm{s} .\) What is its speed at \(B,\) where \(x_{B}=-2.0 \mathrm{m} ?\)

In a common physics demonstration, a bowling ball is suspended from the ceiling by a rope. The professor pulls the ball away from its equilibrium position and holds it adjacent to his nose, as shown below. He releases the ball so that it swings directly away from him. Does he get struck by the ball on its return swing? What is he trying to show in this demonstration?

Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of gasoline after a short distance (see below). The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred in this series of events.
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