Pendulum RotationA simple pendulumCan an integral of a function that is not well behaved be finite?Lagrangian...
Slow moving projectiles from a hand-held weapon - how do they reach the target?
I am on the US no-fly list. What can I do in order to be allowed on flights which go through US airspace?
Why avoid shared user accounts?
Program that converts a number to a letter of the alphabet
integral inequality of length of curve
En Passant For Beginners
Am I a Rude Number?
What to do when being responsible for data protection in your lab, yet advice is ignored?
Why did Jodrell Bank assist the Soviet Union to collect data from their spacecraft in the mid 1960's?
What is better: yes / no radio, or simple checkbox?
Can you earn endless XP using a Flameskull and its self-revival feature?
Closed form for these polynomials?
What kind of hardware implements Fourier transform?
Overfitting and Underfitting
What is the etymology of the kanji 食?
Are there any outlying considerations if I treat donning a shield as an object interaction during the first round of combat?
Why can a 352GB NumPy ndarray be used on an 8GB memory macOS computer?
Pendulum Rotation
Tikzing a circled star
How to remove trailing forward slash
The vanishing of sum of coefficients: symmetric polynomials
What makes the Forgotten Realms "forgotten"?
Getting a UK passport renewed when you have dual nationality and a different name in your second country?
Why is button three on trumpet almost never used alone?
Pendulum Rotation
A simple pendulumCan an integral of a function that is not well behaved be finite?Lagrangian of bead on a rotating hoopEuler-Lagrange - circle coneInclined plane - euler-lagrangeAngular Velocity calculationQuestion on inclined plane with velocity of projection $u$ both up and down plane.Finding the Velocity of a Particle after an ImpactWhat is the equation of the trajectory from the second bounce?Physics answers obtained through two different mathematical approaches contradict.
$begingroup$
A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.
I am trying to use this expression to find the maximum velocity of the bob.
$$omega^2_f=2 alphatriangletheta$$
I get the following expression:
$$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$
Is this expression correct to solve the question?
mathematical-physics
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
$endgroup$
add a comment |
$begingroup$
A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.
I am trying to use this expression to find the maximum velocity of the bob.
$$omega^2_f=2 alphatriangletheta$$
I get the following expression:
$$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$
Is this expression correct to solve the question?
mathematical-physics
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
$endgroup$
add a comment |
$begingroup$
A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.
I am trying to use this expression to find the maximum velocity of the bob.
$$omega^2_f=2 alphatriangletheta$$
I get the following expression:
$$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$
Is this expression correct to solve the question?
mathematical-physics
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
$endgroup$
A simple pendulum has a bob with a mass of .50 kg. The cord has a length of 1.5 m, and the bob is displaced $20^circ$.
I am trying to use this expression to find the maximum velocity of the bob.
$$omega^2_f=2 alphatriangletheta$$
I get the following expression:
$$frac{v^2}{L^2}=2Lfrac{mgsintheta}{I}triangletheta$$
Is this expression correct to solve the question?
mathematical-physics
mathematical-physics
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
edited 1 hour ago
EnlightenedFunky
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
asked 1 hour ago
EnlightenedFunkyEnlightenedFunky
81211022
81211022
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
Notice that the velocity is independent of mass
answered 1 hour ago
AndreiAndrei
12.5k21128
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
|
show 1 more comment
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3133059%2fpendulum-rotation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
Notice that the velocity is independent of mass
answered 1 hour ago
AndreiAndrei
12.5k21128
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
|
show 1 more comment
$begingroup$
First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
Notice that the velocity is independent of mass
answered 1 hour ago
AndreiAndrei
12.5k21128
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
|
show 1 more comment
$begingroup$
First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
Notice that the velocity is independent of mass
answered 1 hour ago
AndreiAndrei
12.5k21128
$endgroup$
First of all, you should explain what your parameters are. But you seem to over complicate the problem. Write conservation of energy. At the bottom of the trajectory, we choose the potential energy to be $0$. Then you have only kinetic energy $frac 12 mv^2$. At the top of the trajectory, the bob is at rest, so kinetic energy is zero, and you have only potential energy. The height of the bob is $L(1-costheta)$, so $$mgL(1-costheta)=frac 12 mv^2$$
Notice that the velocity is independent of mass
answered 1 hour ago
AndreiAndrei
12.5k21128
answered 1 hour ago
AndreiAndrei
12.5k21128
answered 1 hour ago
AndreiAndrei
12.5k21128
answered 1 hour ago
AndreiAndrei
12.5k21128
12.5k21128
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
|
show 1 more comment
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
You did not explain what the $alpha$ or $I$ or $omega_f$ are. I guessed that you they represent angular acceleration, moment of inertia, and angular velocity. However, the formula that you wrote cannot be used in this case. That formula is valid for constant $alpha$, which is not the case here. The torque depends on the angle between gravity and the cord.
$endgroup$
– Andrei
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
1 hour ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
57 mins ago
|
show 1 more comment
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=frac{gsintheta}{l}$, which depends on $theta$
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
answered 1 hour ago
Yuzheng LinYuzheng Lin
411
411
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yuzheng Lin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3133059%2fpendulum-rotation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
