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Pendulum Rotation
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$begingroup$
A simple pendulum has a bob with a mass of $0.50$ 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.
$$omega^2_f = 2 alphaDeltatheta$$
I get the following expression:
$$frac{v^2}{L^2} = 2Lfrac{mgsintheta}{I}Deltatheta$$
Is this expression correct to solve the question?
mathematical-physics
edited 4 hours ago
Rócherz
2,8062721
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
$endgroup$
add a comment |
$begingroup$
A simple pendulum has a bob with a mass of $0.50$ 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.
$$omega^2_f = 2 alphaDeltatheta$$
I get the following expression:
$$frac{v^2}{L^2} = 2Lfrac{mgsintheta}{I}Deltatheta$$
Is this expression correct to solve the question?
mathematical-physics
edited 4 hours ago
Rócherz
2,8062721
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
$endgroup$
add a comment |
$begingroup$
A simple pendulum has a bob with a mass of $0.50$ 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.
$$omega^2_f = 2 alphaDeltatheta$$
I get the following expression:
$$frac{v^2}{L^2} = 2Lfrac{mgsintheta}{I}Deltatheta$$
Is this expression correct to solve the question?
mathematical-physics
edited 4 hours ago
Rócherz
2,8062721
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
$endgroup$
A simple pendulum has a bob with a mass of $0.50$ 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.
$$omega^2_f = 2 alphaDeltatheta$$
I get the following expression:
$$frac{v^2}{L^2} = 2Lfrac{mgsintheta}{I}Deltatheta$$
Is this expression correct to solve the question?
mathematical-physics
mathematical-physics
edited 4 hours ago
Rócherz
2,8062721
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
edited 4 hours ago
Rócherz
2,8062721
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
edited 4 hours ago
Rócherz
2,8062721
edited 4 hours ago
Rócherz
2,8062721
edited 4 hours ago
Rócherz
2,8062721
2,8062721
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
asked 12 hours ago
EnlightenedFunkyEnlightenedFunky
82211022
82211022
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 11 hours ago
AndreiAndrei
12.6k21128
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
4
$begingroup$
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
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours 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 11 hours 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 |
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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 11 hours ago
AndreiAndrei
12.6k21128
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
4
$begingroup$
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
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours 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 11 hours ago
AndreiAndrei
12.6k21128
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
4
$begingroup$
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
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours 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 11 hours ago
AndreiAndrei
12.6k21128
$endgroup$
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 11 hours ago
AndreiAndrei
12.6k21128
answered 11 hours ago
AndreiAndrei
12.6k21128
answered 11 hours ago
AndreiAndrei
12.6k21128
answered 11 hours ago
AndreiAndrei
12.6k21128
12.6k21128
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
4
$begingroup$
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
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours ago
|
show 1 more comment
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
4
$begingroup$
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
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours ago
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
11 hours ago
4
4
$begingroup$
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
11 hours 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
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours ago
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
11 hours 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 11 hours 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 11 hours 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 11 hours 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 11 hours 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 11 hours 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 11 hours ago
Yuzheng LinYuzheng Lin
411
answered 11 hours 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 |
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