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Chapter 10: Problem 20

Pendulum shows a pendulum of length \(L\). Its bob (which effectively has all the mass) has speed \(v_{1}\) when the cord makes an angle \(\theta_{1}\) with the vertical. (a) Derive an expression for the speed of the bob when it is in its lowest position. What is the least value that \(v_{1}\) 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 \(\theta_{1}\) is increased by a few degrees?

Short Answer

Expert verified
The speed in the lowest position is \ v_2 = \sqrt{v_1^2 + 2gL(1 - \cos\theta_1)} \. Minimal \ v_1 \ for horizontal is \ \sqrt{2gL(\cos\theta_1)} \ and for vertical is \ \sqrt{5gL(1 - \cos\theta_1)} \. All increase with \ \theta_1 \.

Step by step solution

01

Identify Energy Conservation

The system conserves mechanical energy. At any point, kinetic energy (KE) and potential energy (PE) sum up to a constant value.
02

Write Expression for Initial State

At angle \( \theta_1 \) with speed \( v_1 \), the energies are given by:\[ KE_1 = \frac{1}{2}mv_1^2 \] \[ PE_1 = mgL(1 - \cos\theta_1) \]
03

Write Expression for Lowest Position

At the lowest position (vertical), all potential energy converts to kinetic energy:\[ KE_2 = \frac{1}{2}mv_2^2 \] \[ PE_2 = 0 \]
04

Equate Energies and Simplify

Using energy conservation \(KE_1 + PE_1 = KE_2 \):\[ \frac{1}{2}mv_1^2 + mgL(1 - \cos\theta_1) = \frac{1}{2}mv_2^2 \]Solve for \ v_2 \[ v_2 = \sqrt{v_1^2 + 2gL(1 - \cos\theta_1)} \]
05

Minimum Speed For Horizontal Swing (b)

For the bob to reach the horizontal position, it must have enough energy to have \( PE_2 = mgL \):\[ \frac{1}{2}mv_{1_{min}}^2 + mgL(1 - \cos\theta_1) = mgL \]Solve for \ v_{1_{min}} \[ v_{1_{min}} = \sqrt{2gL(\cos\theta_1)} \]
06

Minimum Speed For Vertical Swing (c)

For the bob to reach the vertical position on the opposite side, it must have enough energy to have \( PE_2 = 2mgL \):\[ \frac{1}{2}mv_{1_{min}}^2 + mgL(1 - \cos\theta_1) = 2mgL \]Solve for \ v_{1_{min}} \[ v_{1_{min}} = \sqrt{5gL(1 - \cos\theta_1)} \]
07

Effect of Increasing \(\theta_1\) (d)

As \( \theta_1 \) increases, term \( (1 - \cos\theta_1) \) increases. Hence \( v_2 \), \( v_{1_{min}} \) for the horizontal position, and \( v_{1_{min}} \) for the vertical position all increase.

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Key Concepts

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

Energy Conservation
In the motion of a pendulum, energy conservation is a key principle. This means that the total mechanical energy of the system - which is the sum of kinetic energy (KE) and potential energy (PE) - remains constant. As the pendulum swings back and forth, energy is continuously converted between kinetic and potential forms, but the total energy remains the same, assuming no energy is lost to friction or air resistance.
Potential Energy
Potential energy in a pendulum is related to the height of the bob. When the pendulum bob is at a higher position (like at the angle \( \theta_1 \)), it has more potential energy. The formula for potential energy at a height is:
\[ PE = mgL(1 - \cos\theta_1) \]
In this formula, m is the mass of the bob, g is the acceleration due to gravity, and L is the length of the string.
When the pendulum bob is at its lowest point, the potential energy is at its minimum (zero, if we choose this point as our reference height). This shows how height and angle affect the potential energy of the bob.
Kinetic Energy
Kinetic energy is the energy associated with the motion of the pendulum bob. When the bob swings through the lowest point, it has its maximum speed and thus the maximum kinetic energy. The kinetic energy formula for the bob's speed is:
\[ KE = \frac{1}{2}mv^2 \]
At the highest point of the swing, where the bob momentarily stops before reversing direction<, its kinetic energy is zero; all the energy is potential energy. This exchange of kinetic and potential energy keeps the pendulum moving.
Minimum Speed
To understand the minimum speed required for the pendulum bob to reach certain positions, we need to consider the energy required to get there.
For the bob to reach a horizontal position, it must have enough initial kinetic and potential energy to gain the potential energy corresponding to that position:
\[ v_1^{min,horiz} = \sqrt{2gL(1 - \cos\theta_1)} \]
For the bob to reach the vertical position on the opposite side, it needs even more energy:
\[ v_1^{min,vert} = \sqrt{5gL(1 - \cos\theta_1)} \]
These formulas help us determine if the initial speed is sufficient for the pendulum to reach the desired positions.
Effect of Angle on Motion
The angle \( \theta_1 \) at which the pendulum bob is released has a significant impact on its motion. As the release angle increases, the term \( (1 - \cos\theta_1) \) also increases. This means that:
  • The speed at the lowest position \( v_2 \) increases.

  • The minimum initial speed required to reach the horizontal position increases.

  • The minimum initial speed required to reach the vertical position on the opposite side also increases.

Thus, releasing the pendulum from a higher initial angle gives the bob more potential energy to convert into other forms, affecting its motion substantially.

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Most popular questions from this chapter

Stone Rests on Spring shows an \(8.00 \mathrm{~kg}\) stone at rest on a spring. The spring is compressed \(10.0 \mathrm{~cm}\) by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional \(30.0\) \(\mathrm{cm}\) and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?

Block Drawn by Rope A \(3.57 \mathrm{~kg}\) block is drawn at constant speed \(4.06 \mathrm{~m}\) along a horizontal floor by a rope. The force on the block from the rope has a magnitude of \(7.68 \mathrm{~N}\) and is directed \(15.0^{\circ}\) above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

Block Dropped on a Spring A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) 2.5 N/cm (Fig. 10-32). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Ballistic Pendulum A bullet of mass \(10 \mathrm{~g}\) strikes a ballistic pendulum of mass \(2.0 \mathrm{~kg}\). The center of mass of the pendulum rises a vertical distance of \(12 \mathrm{~cm}\). Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Steel Ball and Block A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 10-61). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
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