As Pendulum Swings Back and Forth Potential Energy Is Tranformed Into Kinetic

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Consider the diagram of a pendulum shown below.

11 26 15

As the pendulum swings back and forth, which of the following values is at its maximum when the pendulum is at the bottom of its line of motion?

Possible Answers:

Period of oscillation

Kinetic energy

Tangential acceleration

Frequency of oscillation

Potential energy

Correct answer:

Kinetic energy

Explanation:

For this question, we're presented with a scenario in which a pendulum is swinging back and forth. Thus, we know that this pendulum is an example of simple harmonic motion. As the pendulum swings back and forth, a number of its variables change in a cyclical fashion.

First, let's take a look at potential energy. When the pendulum is at the bottom of its trajectory, its potential energy will also be at its minimum. This is because the height of the mass attached to the pendulum is lowest at this point. We can show this with the following expression, where the term is at a minimum.

U=mgh

Next, let's consider tangential acceleration. When the pendulum reaches its highest point, it will briefly be at rest for a very short instant. At this highest point, the pendulum also has its greatest amount of potential energy. When the pendulum begins to fall, the force due to gravity causes the pendulum to fall. However, it's important to realize that the force of gravity acts on the pendulum's mass in two ways. One way is tangentially, in which the force acts along the direction of the mass's motion. The other way is radially to the mass's direction of motion; in other words, along the pendulum's string. This can be shown with a diagram as follows.

11 26 15 pendulum acceleration

As can be seen in the above diagram, the tangential acceleration is represented by the following expression.

$a=gsin(\Theta )$

Thus, as the pendulum swings to its lowest point, the value of $\Theta$ approaches zero. As it does this, the tangential acceleration also approaches zero.

Both the frequency and the period of the pendulum's harmonic motion is in no way related to the lowest point of the pendulum's path.

Finally, let's consider kinetic energy. We've already noted how the pendulum's potential energy is at a maximum at its highest point. As the pendulum falls to its lowest point, its potential energy is converted into kinetic energy. This is because as the pendulum falls to its lowest point, it speeds up more and more. Thus, at its lowest point, the pendulum has its kinetic energy at a maximum.

Which simple pendulum will have a longer period?

$A:m = 10\textup{ kg}, L = 2\textup{ m}$

$B: m = 4\textup{ kg}, L = 3\textup{ m}$

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

Possible Answers:

B, because it has a smaller mass

A, because it has a shorter length

A, because it has a larger mass

They will have the same period

B, because it has a longer length

Correct answer:

B, because it has a longer length

Explanation:

The expression for the period of a pendulum is:

$T = 2\pi \sqrt{\frac{L}{g}}$

Therefore, the period of a pendulum is proportional to the square root of the length of the pendulum (assuming they are both on earth, or the same planetary body). Thus, the pendulum with the longer length will have the longer period.

A simple pendulum of length $L = 0.6\textup{ m}$ with a block of mass $m = 5\textup{ kg}$ attached has a maximum velocity of $1.2\ \frac{\textup{m}}{\textup{s}^2}$ . What is the maximum height of the block?

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

Correct answer:

$0.07\textup{ m}$

Explanation:

This may come as a surprise, but we don't need to know a single formula concerning pendulums or circular motion to solve this problem. We just have to be able to understand the motion of a pendulum and think about the situation practically. A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the expression for conservation of energy to determine the maximum height of the block.

E = U_i+K_i=U_f+K_f

If we assume that the low point of the pendulum has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy (at least for now. If we get a maximum height that is more than twice the length of the pendulum, we know that it completes full rotations) since the block will be at rest when it reaches its maximum height.

K_i = U_f

Substituting expressions in for each term:

$\frac{1}{2}mv_i^2=mgh_f$

Eliminating mass and rearranging for final height, we get:

$h_f = \frac{v^2}{2g}$

Plugging in our values, we get:

$h_f = \frac{\left ( 1.2\frac{m}{s}\right )^2}{2\left ( 10\frac{m}{s^2}\right )}$

h_f = 0.07m

A simple pendulum with a length of $L = 2\textup{ m}$ has a block of mass $m = 6\textup{ kg}$ attached to the end. If the pendulum is $h=1\textup{ m}$ above its lowest point and rotating downward, what is the instantaneous acceleration of the block?

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

Correct answer:

$5.0\ \frac{\textup{m}}{\textup{s}^2}$

Explanation:

Since we are told that the block is 1m above its lowest point (half way between its low point and horizontal), we can calculate the angle that the pendulum makes with the vertical:

$sin(\theta) = \frac{h}{L}= \frac{1m}{2m} = \frac{1}{2}$

$\theta = 30^{\circ}$

Then we can use the following expression to determine the net force acting on the block in the direction of its motion:

$F_{net}= mgsin(\theta)$

Then we can use Newton's 2nd law to determine the instantaneous acceleration:

$F_{net}=ma = mgsin(\theta)$

Canceling out mass and rearranging for acceleration:

$a = gsin(\theta)$

$a = \left ( 10\frac{m}{s^2}\right )\left ( \frac{1}{2}\right )$

$a = 5\frac{m}{s^2}$

A block of mass $m =8\textup{ kg}$ is attached to a rigid pole of length $L = 2\textup{ m}$ . If the block has a velocity of $4\ \frac{\textup{m}}{\textup{s}}$ as it travels through the horizontal, what is the distance between the blocks lowest and highest points? Neglect the mass of the pole. Neglect air resistance and any frictional forces.

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

Correct answer:

$2.8\textup{ m}$

Explanation:

Since we know the velocity of the block as it travels through the horizontal, we can directly calculate the distance above horizontal that the block reaches using the expression for conservation of energy:

E = U_i + K_i = U_f+K_f

If we assume that horizontal has a height of 0, we can eliminate initial potential energy. We can also eliminate final kinetic energy as the block should be at rest when it reaches its highest point.

K_i = U_f

Plugging in expressions for each variable, we get:

$\frac{1}{2}mv_i^2=mgh_f$

Eliminating mass and rearranging for final height, we get:

$h_f = \frac{v_i^2}{2g}$

Plugging in our values, we get:

$h_f = \frac{\left ( 4\frac{m}{s}\right )^2}{2\left ( 10\frac{m}{s^2}\right )}$

h_f = 0.8m

This is the distance the block reaches above the horizontal, so we need to add this to the distance between horizontal and the blocks lowest point, which is simply the length of the pole:

d = L+h = 2m+0.8m=2.8m

A rigid rod of length $L = 4\textup{ m}$ has a block of mass $m=25\textup{ kg}$ attached to one end and is allowed to rotate as a simple pendulum from the other end. What is the lowest maximum velocity of the block that will allow the block to rotate in complete circles?

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

Correct answer:

$12.6\ \frac{\textup{m}}{\textup{s}}$

Explanation:

The lowest maximum velocity occurs in the scenario where the block is at rest when the pendulum is vertical and pointing upward. Therefore, even the slightest additional movement will result in the pendulum rotating in complete circles With this in mind, we can begin with the expression for conservation of energy:

E = U_i +K_i = U_f+K_f

If we say the initial condition is when the block is at the lowest point of rotation (when it is traveling at its maximum velocity), and assume that point to have a height of 0, we can eliminate initial potential energy. Furthermore, our final state will then be when the block is at the highest point of rotation and at rest. Thus we can eliminate final kinetic energy to get:

K_i = U_f

Plugging in expressions for each variable, we get:

$\frac{1}{2}mv_{max}^2 =mgh_{max}$

Eliminating mass and rearranging for maximum velocity, we get:

$v_{max}=\sqrt{2gh_{max}}$

Where the maximum height is two times the length of the pendulum:

$v_{max}=\sqrt{2\left ( 10\frac{m}{s^2}\right )(8m)}$

$v_{max} = 12.6\frac{m}{s}$

Correct answer:

$4.8^{\circ}$

Explanation:

Since the maximum angle achieved by the pendulum is very small, we can use the follow expression to determine the angle of the pendulum at any time :

$\theta = \theta_{max}cos\left (\sqrt{\frac{g}{L}}t \right )$

Note how we are using the cosine function since the pendulum began at its highest point. We already have all of these values, so we can simply plug and chug:

$\theta = (5^{\circ})cos{\left ( \sqrt{\frac{\left ( 10\frac{m}{s^2}\right )}{4m}}(10s)\right )}$

$\theta = 1.4^{\circ}$

A simple pendulum of length $L = 2.5\textup{ m}$ has a block attached to one end which has a maximum velocity of $12.5\ \frac{\textup{m}}{\textup{s}}$ . What is the minimum velocity of the block?

$g = 10\frac{m}{s^2}$

Correct answer:

$7.5\ \frac{\textup{m}}{\textup{s}}$

Explanation:

The block experiences its maximum velocity when it is at its lowest point of rotation and its minimum velocity at its highest point of rotation. Therefore, we can use the expression for conservation of energy to solve this problem:

E = U_i +K_i = U_f +K_f

If we assume that the initial condition is when the block is at the low point of rotation and assume that that point has a height of 0, then we can eliminate initial potential energy:

K_i = U_f +K_f

Now substituting in expressions for each of these:

$\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2$

Eliminating mass and multiplying both sides of the expression by 2, we get:

v_i^2 = 2gh_f+v_f^2

Then rearranging for final velocity:

$v_f = \sqrt{v_i^2-2gh_f}$

Where the height is simply twice the length of the pendulum:

$v_f = \sqrt{v_i^2-4gl}$

Plugging in our values:

$v_f = \sqrt{\left ( 12.5\frac{m}{s}\right )^2-4\left ( 10\frac{m}{s^2}\right )(2.5m)}$

$v_f = 7.5\frac{m}{s}$

A simple pendulum has a length $L = 1.4\textup{ m}$ has a block of mass $m = 6\textup{ kg}$ attached to one end. If the pendulum is released from rest, what is the maximum centripetal acceleration felt by the block?

$g = 10\ \frac{\textup{m}}{\textup{s}^2}$

Correct answer:

$20\ \frac{\textup{m}}{\textup{s}^2}$

Explanation:

Since we are given the length of the pendulum and told that it begins at rest and in the horizontal position, we can calculate the maximum velocity of the block as it travels through its lowest point using the expression for conservation of energy:

E = U_i +K_i = U_f + K_f

We can eliminate initial kinetic energy since the pendulum begins at rest. We can also eliminate final potential energy if we assume that the height at the lowest point of rotation is equal to 0.

U_i = K_f

Substituting in expressions for each of these:

$mgh_i = \frac{1}{2}mv_f^2$

Where the initial height is just the length of the pendulum:

$mgL= \frac{1}{2}mv_f^2$

Rearranging for final velocity, we get:

$v_f = \sqrt{2gL}$

We can then calculate centripetal acceleration from this:

$a_c = \frac{v^2}{r}$

Where the radius is the length of the pendulum:

$a_c = \frac{v^2}{L} = \frac{2gL}{L} = 2g$

$a_c = 20\frac{m}{s^2}$

A simple pendulum with a length $L = 4\textup{ m}$ has a block attached to one end that has maximum velocity of $v = 8\ \frac{\textup{m}}{\textup{s}}$ . At what angle to the vertical does the block have a velocity of $4\ \frac{\textup{m}}{\textup{s}^2}$ ?

$g=10\ \frac{\textup{m}}{\textup{s}^2}$

Correct answer:

$66.4^{\circ}$

Explanation:

We can begin with the expression for conservation of energy to solve this problem:

E = U_i+K_i = U_f +K_f

The block will achieve its maximum velocity at the lowest point of rotation. If we say that this point has a height of 0, we can eliminate initial potential:

K_i = U_f +K_f

Plugging in expressions for each of these:

$\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2$

Multiplying both sides of the expression by $\frac{2}{m}$ , we get:

v_i^2 = 2gh_f+v_f^2

Now let's say that the final condition is when the block has a velocity of $4\frac{m}{s}$ . Rearranging for the final height, we get:

$h_f= \frac{v_i^2-v_f^2}{2g}$

Plugging in our values:

$h_f = \frac{\left ( 8\frac{m}{s}\right )^2-\left ( 4\frac{m}{s}\right )^2}{2\left ( 10\frac{m}{s^2}\right )}$

h_f = 2.4m

This is the height that the block is above our reference point when it reaches the desired speed. From here, we can develop an expression for the height of the block as a function of the angle the block makes with the vertical. First, let's begin with a function that tells us how far below the block is from the horizontal:

$d = Lcos(\theta)$

If this does not make sense, draw it out. d is the distance below the horizontal, L is the length of the pendulum, and theta is the angle between the pendulum and the horizontal.

Moving on, we can take this expression to develop one that tells us the height of the block:

$h = L-Lcos(\theta) = L(1-cos(\theta))$

Rearranging for angle:

$1-cos(\theta)=\frac{h}{L}$

$\theta = cos^{-1}\left ( 1-\frac{h}{L}\right )$

Now plugging in our values:

$\theta = cos^{-1}\left ( 1-\frac{2.4m}{4m}\right )$

$\theta = 66.4^{\circ}$

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As Pendulum Swings Back and Forth Potential Energy Is Tranformed Into Kinetic

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As Pendulum Swings Back and Forth Potential Energy Is Tranformed Into Kinetic

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