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Chapter 8: Problem 21

Shows a pendulum of length \(L=1.25 \mathrm{~m}\). Its bob (which effectively has all the mass) has speed \(v_{0}\) when the cord makes an angle \(\theta_{0}=40.0^{\circ}\) with the vertical. (a) What is the speed of the bob when it is in its lowest position if \(v_{0}=8.00 \mathrm{~m} / \mathrm{s} ?\) What is the least value that \(v_{0}\) can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if \(\theta_{0}\) is increased by a few degrees?

Short Answer

Expert verified
(a) Apply conservation of energy to find speed at lowest point. (b) Minimum speed \( v_0 = \sqrt{2gL\cos \theta_0} \). (c) Minimum speed \( v_0 = \sqrt{5gL \cos \theta_0} \). (d) Answers decrease with increased \( \theta_0 \).

Step by step solution

01

Understand the Problem

We have a pendulum of length \( L = 1.25 \, \text{m} \). The initial speed \( v_0 = 8.00 \, \text{m/s} \) when the pendulum makes an angle \( \theta_0 = 40.0^\circ \) with the vertical. We need to find the speeds at different positions and the least initial speed for specific movements.
02

Calculate Speed at the Lowest Position

Use conservation of mechanical energy. At the highest point: \[ E_i = \frac{1}{2}mv_0^2 + mgL(1 - \cos \theta_0) \]At the lowest point (\(\theta = 0\)): \[ E_f = \frac{1}{2}mv^2 \]Equate initial and final energy to solve for \( v \):\[ \frac{1}{2}mv_0^2 + mgL(1 - \cos \theta_0) = \frac{1}{2}mv^2 \]Solve for \( v \):\[ v = \sqrt{v_0^2 + 2gL(1 - \cos \theta_0)} \]Substitute \( v_0 = 8.00 \, \text{m/s}, \theta_0 = 40^\circ, g = 9.81 \, \text{m/s}^2 \) into the equation and calculate \( v \).
03

Calculate Minimum Initial Speed for Horizontal Position

To reach the horizontal, the pendulum's bob needs to have enough kinetic energy to convert to potential energy at 90掳 (\(L\)).The required energy at horizontal: \[ mgL \]Equate initial kinetic and gravitational energy changes:\[ \frac{1}{2}mv_0^2 + mgL(1 - \cos \theta_0) = mgL \]Solve for \( v_0 \):\[ v_0 = \sqrt{2gL\cos \theta_0} \]Calculate \( v_0 \) using given values.
04

Calculate Minimum Initial Speed for Vertical Position

To reach the vertical, the bob needs to have energy to reach a height of \( 2L \).The required energy at completely vertical:\[ 2mgL \]Equate initial and potential energy changes:\[ \frac{1}{2}mv_0^2 + mgL(1 - \cos \theta_0) = 2mgL \]Solve for \( v_0 \):\[ v_0 = \sqrt{5gL \cos \theta_0} \]Calculate \( v_0 \) using the values given.
05

Analyze Variation with Increased \( \theta_0 \)

For parts (b) and (c), analyze how the equations for \( v_0 \) depend on \( \theta_0 \). The cosine function in both equations implies that as \( \theta_0 \) increases, \( \cos \theta_0 \) decreases, thereby decreasing the value of \( v_0 \). Hence, increasing \( \theta_0 \) slightly would decrease the least initial speed required to reach those positions.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a crucial concept in physics, especially when dealing with pendulum motion. In simple terms, this principle states that the total energy of an isolated system remains constant over time, regardless of the processes happening within the system.
This means the energy can neither be created nor destroyed; it can only be transformed from one form to another. For a pendulum, this involves a continuous exchange between gravitational potential energy and kinetic energy.
  • At its highest point, the pendulum has maximum potential energy and zero kinetic energy.
  • As it swings downwards, potential energy is converted into kinetic energy, increasing the pendulum's speed.
  • At the lowest point, kinetic energy is at its peak, while potential energy is at its minimum.
This continuous transformation between kinetic and potential energy allows the pendulum to keep swinging back and forth, demonstrating conservation of energy in action.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. When applied to pendulum motion, kinematics helps us understand important aspects such as speed, velocity, and displacement.
For a pendulum, these variables change as the pendulum swings back and forth.
Consider these kinematic notions in pendulum motion:
  • Displacement: The pendulum's bob moves in an arc, and its displacement can be measured by the angle it makes with the vertical.
  • Velocity: The velocity varies as the pendulum moves, reaching its maximum at the lowest point of the swing.
  • Acceleration: The greatest acceleration occurs at the tips of the swing, when the direction changes.
The kinematic equations that describe pendulum motion take into account these changes over time, allowing us to predict positions and speeds during the pendulum's trajectory.
Gravitational Potential Energy
Gravitational potential energy refers to the energy that an object possesses because of its position in a gravitational field. In the context of a pendulum, this energy is attributed to the height of the pendulum above its lowest point.
The formula for gravitational potential energy is given by:\[ U = mgh \]where:
  • U denotes the gravitational potential energy.
  • m is the mass of the pendulum's bob.
  • g stands for the acceleration due to gravity.
  • h represents the height of the bob relative to the lowest point of the swing.
As the pendulum rises, the height (and thus the gravitational potential energy) increases. Conversely, as it descends, the energy converts into kinetic energy, which is observable as motion. Understanding this energy conversion is key to comprehending pendulum dynamics.
Mechanical Energy
Mechanical energy encompasses both the kinetic and potential energy of an object in motion. For a pendulum, mechanical energy is the sum of its gravitational potential energy and its kinetic energy at any point during its swing.
Mechanical energy is critical to solving pendulum problems because it remains constant when considering an ideal pendulum with no energy losses (such as air resistance or friction).
  • Kinetic Energy (KE): Calculated by the equation \( KE = \frac{1}{2}mv^2 \), it increases as the pendulum moves towards the bottom.
  • Gravitational Potential Energy (GPE): Calculated by \( GPE = mgh \), it is at its highest when the pendulum is at its peak points.
  • Using the conservation of mechanical energy, one can predict velocities or calculate minimum initial speeds for achieving certain positions.
Mechanical energy provides the framework for analyzing pendulum motion and the potential and kinetic energy transformations involved.

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Most popular questions from this chapter

A uniform cord of length \(25 \mathrm{~cm}\) and mass \(15 \mathrm{~g}\) is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation?

A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is \(0.10 .\) (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s}\), what is her speed at the bottom?

A block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance \(d\). The block's initial speed \(v_{0}\) is \(6.0 \mathrm{~m} / \mathrm{s}\), the height difference \(h\) is \(1.1 \mathrm{~m}\), and \(\mu_{k}\) is \(0.60\). Find \(d\).

A block of mass \(m=2.5 \mathrm{~kg}\) slides head on into a spring of spring constant \(k=320 \mathrm{~N} / \mathrm{m}\). When the block stops, it has compressed the spring by \(7.5 \mathrm{~cm}\). The coefficient of kinetic friction between block and floor is \(0.25\). While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block's speed just as it reaches the spring?

A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. 8-35) and leaves the ramp at angle \(\theta=28^{\circ} .\) Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?
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