Pendulum Rotation
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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:
$$fracv^2L^2 = 2LfracmgsinthetaIDeltatheta$$
Is this expression correct to solve the question?
calculus
edited Mar 6 at 19:41
Andrews
1,2812422
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
$endgroup$
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:
$$fracv^2L^2 = 2LfracmgsinthetaIDeltatheta$$
Is this expression correct to solve the question?
calculus
edited Mar 6 at 19:41
Andrews
1,2812422
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
$endgroup$
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:
$$fracv^2L^2 = 2LfracmgsinthetaIDeltatheta$$
Is this expression correct to solve the question?
calculus
edited Mar 6 at 19:41
Andrews
1,2812422
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
$endgroup$
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:
$$fracv^2L^2 = 2LfracmgsinthetaIDeltatheta$$
Is this expression correct to solve the question?
calculus
calculus
edited Mar 6 at 19:41
Andrews
1,2812422
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
edited Mar 6 at 19:41
Andrews
1,2812422
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
edited Mar 6 at 19:41
Andrews
1,2812422
edited Mar 6 at 19:41
Andrews
1,2812422
edited Mar 6 at 19:41
Andrews
1,2812422
1,2812422
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
asked Mar 3 at 1:40
EnlightenedFunkyEnlightenedFunky
83811022
83811022
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 Mar 3 at 2:05
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
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
Mar 3 at 2:15
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
|
show 3 more comments
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=fracgsinthetal$, which depends on $theta$
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
$endgroup$
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 Mar 3 at 2:05
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
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
Mar 3 at 2:15
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
|
show 3 more comments
$begingroup$
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 Mar 3 at 2:05
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
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
Mar 3 at 2:15
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
|
show 3 more comments
$begingroup$
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 Mar 3 at 2:05
AndreiAndrei
13.2k21230
$endgroup$
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 Mar 3 at 2:05
AndreiAndrei
13.2k21230
answered Mar 3 at 2:05
AndreiAndrei
13.2k21230
answered Mar 3 at 2:05
AndreiAndrei
13.2k21230
answered Mar 3 at 2:05
AndreiAndrei
13.2k21230
13.2k21230
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
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
Mar 3 at 2:15
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
|
show 3 more comments
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
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
Mar 3 at 2:15
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
But, I would like to solve this particular manner the professor solved it that way already.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:10
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
$begingroup$
Secondly, everything was given to you.in the detail.
$endgroup$
– EnlightenedFunky
Mar 3 at 2:11
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
Mar 3 at 2:15
$begingroup$
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
Mar 3 at 2:15
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
Isn't the force is $mgsintheta$
$endgroup$
– EnlightenedFunky
Mar 3 at 2:19
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
$begingroup$
No, the force is $mg$ always pointing downwards. The torque is $mgLsintheta$
$endgroup$
– Andrei
Mar 3 at 2:22
|
show 3 more comments
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=fracgsinthetal$, which depends on $theta$
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
$endgroup$
add a comment |
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=fracgsinthetal$, which depends on $theta$
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
$endgroup$
add a comment |
$begingroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=fracgsinthetal$, which depends on $theta$
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
$endgroup$
The equation only holds when angular acceleration $alpha$ is a constant. However in this case $alpha=fracgsinthetal$, which depends on $theta$
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
answered Mar 3 at 2:18
Yuzheng LinYuzheng Lin
512
512
add a comment |
add a comment |
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