Pendulum Rotation
$begingroup$
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
59 mins 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
54 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 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 51 mins 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
59 mins 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
54 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 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
59 mins 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
54 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 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
59 mins 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
54 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 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
59 mins 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
54 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 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
59 mins ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
59 mins 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
54 mins 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
54 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
51 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 mins ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
47 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 51 mins 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 51 mins 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 51 mins 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 51 mins 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 51 mins 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 51 mins ago
Yuzheng LinYuzheng Lin
411
answered 51 mins 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
