Figure 8-34 shows a pendulum of length L=1.25 m. Its bob (which effectively has all the mass) h

step 1 All right, energy conservation problem here. We first look at how the pendulum is positioned. So

Figure 8-34 shows a pendulum of length L=1.25 m. Its bob...

Figure $8-34$ 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?

Figure $8-34$ 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?

Key Concepts

Critical Conditions for Circular Motion

When a pendulum or any object is moving in a circular path, there is a minimum speed required at certain points to maintain tension in the string and prevent slack. This concept involves ensuring that the centripetal force, necessary for circular motion, is provided by gravity and the tension in the cord. For positions like the top of the swing (or other specific positions like horizontal or vertical), determining the least initial speed is crucial for ensuring that the pendulum continues its motion without the cord going slack.

Pendulum Kinematics and Geometry

The geometry of a pendulum’s path, including the relationship between the angle of displacement and the vertical height gain, is essential for connecting angular measurements to changes in potential energy. By using trigonometric relations, one can determine how far above the lowest point the pendulum is when at a certain angle, which then feeds into the energy conservation calculations. This geometric insight is key to analyzing the different parts of the question concerning various swing positions.

Gravitational Potential Energy

Gravitational potential energy is the energy an object has due to its position in a gravitational field, and it is defined as mgh (where h is the height above a reference point). In pendulum motion, as the bob swings upward, its kinetic energy is converted into potential energy, and vice versa when it swings downward. Understanding the change in height as the pendulum swings allows one to calculate the corresponding change in energy, which is crucial for finding the bob's speed at various positions.

Conservation of Mechanical Energy

This concept states that in the absence of non-conservative forces (like friction), the total mechanical energy of a system (the sum of kinetic and potential energies) remains constant. For a pendulum, this means that the kinetic energy at one point plus the gravitational potential energy at that same point is equal to the kinetic and potential energy at any other point in its swing. This principle is key for determining the speed at different positions of the pendulum by equating energy at those positions.

Transcript

00:01 All right, energy conservation problem here.

00:03 We first look at how the pendulum is positioned.

00:08 So we have a picture where this is the length of the pendulum.

00:14 And it is displaced by an angle theta not.

00:20 And so it's displaced by an angle theta not.

00:24 Therefore, when it comes up to height h, it goes to so l.

00:33 Goes to l cosine theta not.

00:37 Right.

00:38 So for part a, you have initial kinetic plus potential energy is equal to final kinetic plus potential energy, which means you have one half m times initial velocity squared plus mg times change in height.

00:57 And change in height is l -consign is l -minus l -cosine.

01:01 So it's l times 1 minus cosine theta not.

01:05 That's the delta h there.

01:10 And that's equal to 1 half m v squared v being the final velocity we're solving for.

01:17 This is 0.

01:20 The ms cancel out here.

01:22 Therefore you have, therefore you find a velocity of a velocity expression of inside of a square you have initial velocity squared plus 2 times 2 times l times 1 minus cosine theta not.

01:42 You plug in values now and you get 8 meters per second 1d2 plus 2 times 9 .8 meters per second squared times l which is 1 .25 meters times 1 minus cosine 40 degrees giving you a velocity of 8 .3 .3...

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