Mixing
Mechanical vibration: single degree of freedom model of wheel mounted on a spring
$begingroup$
I think the author of my book has this solution wrong and would like some feedback on my thoughts (see Example 1.4.1).
The solution states that the total energy of the system is represented by the spring, rotation of the wheel, and translation of the wheel, therefore constant total energy E is given by
begin{align} E = (1/2)mv^2 + (1/2)Jomega^2 + (1/2)kx^2 end{align}
where $J$ is the inertia of the wheel, equal to $mr^2$, and omega is the angular velocity equal to $v/r$.
I think in this scenario the translational energy and rotational energy are the same quantity -- that is, there cannot be rotation without translation (since he defined "no slipping occurs and no energy is lost at contact"), and vice-versa; or, any time the object is rotating the energy of rotation will be consumed to produce translation, and vice-versa. Thus the equation should be written
begin{align} E = (1/2)mv^2 + (1/2)kx^2 end{align}
And the maximum kinetic energy will be
begin{align} T_{max} = (1/2)m(omega_n)^2A^2 end{align}
yielding
begin{align} omega_n^2 = k/m end{align}
Thank you in advance for all responses. This is my first time posting on stack exchange.
proof-verification physics classical-mechanics
edited Feb 1 at 4:43
Andrei
13.5k21230
asked Jan 31 at 16:40
LespiegleLespiegle
1
$endgroup$
|
show 3 more comments
$begingroup$
I think the author of my book has this solution wrong and would like some feedback on my thoughts (see Example 1.4.1).
The solution states that the total energy of the system is represented by the spring, rotation of the wheel, and translation of the wheel, therefore constant total energy E is given by
begin{align} E = (1/2)mv^2 + (1/2)Jomega^2 + (1/2)kx^2 end{align}
where $J$ is the inertia of the wheel, equal to $mr^2$, and omega is the angular velocity equal to $v/r$.
I think in this scenario the translational energy and rotational energy are the same quantity -- that is, there cannot be rotation without translation (since he defined "no slipping occurs and no energy is lost at contact"), and vice-versa; or, any time the object is rotating the energy of rotation will be consumed to produce translation, and vice-versa. Thus the equation should be written
begin{align} E = (1/2)mv^2 + (1/2)kx^2 end{align}
And the maximum kinetic energy will be
begin{align} T_{max} = (1/2)m(omega_n)^2A^2 end{align}
yielding
begin{align} omega_n^2 = k/m end{align}
Thank you in advance for all responses. This is my first time posting on stack exchange.
proof-verification physics classical-mechanics
edited Feb 1 at 4:43
Andrei
13.5k21230
asked Jan 31 at 16:40
LespiegleLespiegle
1
$endgroup$
$begingroup$
How can the translational and rotational energy be the same quantity???
$endgroup$
– copper.hat
Jan 31 at 16:43
$begingroup$
The same way potential and kinetic energy are the same when one is completely transformed into the other. What I mean to say is, that if there is a rotational energy of E, all of that energy is going to be used to translate the wheel since it can't be lost to friction and the wheel doesn't slip. Then the rotational energy E will go to 0, and the translational energy which was 0, because the wheel wasn't moving before, is now E.
$endgroup$
– Lespiegle
Jan 31 at 18:41
$begingroup$
I understand the dynamics, but you can't just drop the rotational energy from the equation?
$endgroup$
– copper.hat
Jan 31 at 18:57
$begingroup$
I don't think you do. Let me clarify what I typed earlier. If there is no energy lost to heat then all of the energy created by friction has to be transformed into translational energy or rotational energy. Do you agree with this? Second, the wheel cannot slip. That means that it cannot rotate without creating friction. Therefore it cannot rotate without translating. Do you also agree with this? This is the connection I'm trying to make. I'm not dropping the rotational energy, I'm saying that all of it becomes frictional energy, and then that friction energy translates the wheel.
$endgroup$
– Lespiegle
Jan 31 at 21:15
$begingroup$
(I ran out of room in the last comment). Therefore, one can write either the rotational kinetic energy OR the translational kinetic energy and it will account for both of them simultaneously. Writing both of them would likewise be wrong because you would be counting the kinetic energy of the wheel twice.
$endgroup$
– Lespiegle
Jan 31 at 21:17
|
show 3 more comments
$begingroup$
I think the author of my book has this solution wrong and would like some feedback on my thoughts (see Example 1.4.1).
The solution states that the total energy of the system is represented by the spring, rotation of the wheel, and translation of the wheel, therefore constant total energy E is given by
begin{align} E = (1/2)mv^2 + (1/2)Jomega^2 + (1/2)kx^2 end{align}
where $J$ is the inertia of the wheel, equal to $mr^2$, and omega is the angular velocity equal to $v/r$.
I think in this scenario the translational energy and rotational energy are the same quantity -- that is, there cannot be rotation without translation (since he defined "no slipping occurs and no energy is lost at contact"), and vice-versa; or, any time the object is rotating the energy of rotation will be consumed to produce translation, and vice-versa. Thus the equation should be written
begin{align} E = (1/2)mv^2 + (1/2)kx^2 end{align}
And the maximum kinetic energy will be
begin{align} T_{max} = (1/2)m(omega_n)^2A^2 end{align}
yielding
begin{align} omega_n^2 = k/m end{align}
Thank you in advance for all responses. This is my first time posting on stack exchange.
proof-verification physics classical-mechanics
edited Feb 1 at 4:43
Andrei
13.5k21230
asked Jan 31 at 16:40
LespiegleLespiegle
1
$endgroup$
I think the author of my book has this solution wrong and would like some feedback on my thoughts (see Example 1.4.1).
The solution states that the total energy of the system is represented by the spring, rotation of the wheel, and translation of the wheel, therefore constant total energy E is given by
begin{align} E = (1/2)mv^2 + (1/2)Jomega^2 + (1/2)kx^2 end{align}
where $J$ is the inertia of the wheel, equal to $mr^2$, and omega is the angular velocity equal to $v/r$.
I think in this scenario the translational energy and rotational energy are the same quantity -- that is, there cannot be rotation without translation (since he defined "no slipping occurs and no energy is lost at contact"), and vice-versa; or, any time the object is rotating the energy of rotation will be consumed to produce translation, and vice-versa. Thus the equation should be written
begin{align} E = (1/2)mv^2 + (1/2)kx^2 end{align}
And the maximum kinetic energy will be
begin{align} T_{max} = (1/2)m(omega_n)^2A^2 end{align}
yielding
begin{align} omega_n^2 = k/m end{align}
Thank you in advance for all responses. This is my first time posting on stack exchange.
proof-verification physics classical-mechanics
proof-verification physics classical-mechanics
edited Feb 1 at 4:43
Andrei
13.5k21230
asked Jan 31 at 16:40
LespiegleLespiegle
1
edited Feb 1 at 4:43
Andrei
13.5k21230
asked Jan 31 at 16:40
LespiegleLespiegle
1
edited Feb 1 at 4:43
Andrei
13.5k21230
edited Feb 1 at 4:43
Andrei
13.5k21230
edited Feb 1 at 4:43
Andrei
13.5k21230
13.5k21230
asked Jan 31 at 16:40
LespiegleLespiegle
1
asked Jan 31 at 16:40
LespiegleLespiegle
1
asked Jan 31 at 16:40
LespiegleLespiegle
1
1
$begingroup$
How can the translational and rotational energy be the same quantity???
$endgroup$
– copper.hat
Jan 31 at 16:43
$begingroup$
The same way potential and kinetic energy are the same when one is completely transformed into the other. What I mean to say is, that if there is a rotational energy of E, all of that energy is going to be used to translate the wheel since it can't be lost to friction and the wheel doesn't slip. Then the rotational energy E will go to 0, and the translational energy which was 0, because the wheel wasn't moving before, is now E.
$endgroup$
– Lespiegle
Jan 31 at 18:41
$begingroup$
I understand the dynamics, but you can't just drop the rotational energy from the equation?
$endgroup$
– copper.hat
Jan 31 at 18:57
$begingroup$
I don't think you do. Let me clarify what I typed earlier. If there is no energy lost to heat then all of the energy created by friction has to be transformed into translational energy or rotational energy. Do you agree with this? Second, the wheel cannot slip. That means that it cannot rotate without creating friction. Therefore it cannot rotate without translating. Do you also agree with this? This is the connection I'm trying to make. I'm not dropping the rotational energy, I'm saying that all of it becomes frictional energy, and then that friction energy translates the wheel.
$endgroup$
– Lespiegle
Jan 31 at 21:15
$begingroup$
(I ran out of room in the last comment). Therefore, one can write either the rotational kinetic energy OR the translational kinetic energy and it will account for both of them simultaneously. Writing both of them would likewise be wrong because you would be counting the kinetic energy of the wheel twice.
$endgroup$
– Lespiegle
Jan 31 at 21:17
|
show 3 more comments
$begingroup$
How can the translational and rotational energy be the same quantity???
$endgroup$
– copper.hat
Jan 31 at 16:43
$begingroup$
The same way potential and kinetic energy are the same when one is completely transformed into the other. What I mean to say is, that if there is a rotational energy of E, all of that energy is going to be used to translate the wheel since it can't be lost to friction and the wheel doesn't slip. Then the rotational energy E will go to 0, and the translational energy which was 0, because the wheel wasn't moving before, is now E.
$endgroup$
– Lespiegle
Jan 31 at 18:41
$begingroup$
I understand the dynamics, but you can't just drop the rotational energy from the equation?
$endgroup$
– copper.hat
Jan 31 at 18:57
$begingroup$
I don't think you do. Let me clarify what I typed earlier. If there is no energy lost to heat then all of the energy created by friction has to be transformed into translational energy or rotational energy. Do you agree with this? Second, the wheel cannot slip. That means that it cannot rotate without creating friction. Therefore it cannot rotate without translating. Do you also agree with this? This is the connection I'm trying to make. I'm not dropping the rotational energy, I'm saying that all of it becomes frictional energy, and then that friction energy translates the wheel.
$endgroup$
– Lespiegle
Jan 31 at 21:15
$begingroup$
(I ran out of room in the last comment). Therefore, one can write either the rotational kinetic energy OR the translational kinetic energy and it will account for both of them simultaneously. Writing both of them would likewise be wrong because you would be counting the kinetic energy of the wheel twice.
$endgroup$
– Lespiegle
Jan 31 at 21:17
$begingroup$
How can the translational and rotational energy be the same quantity???
$endgroup$
– copper.hat
Jan 31 at 16:43
$begingroup$
How can the translational and rotational energy be the same quantity???
$endgroup$
– copper.hat
Jan 31 at 16:43
$begingroup$
The same way potential and kinetic energy are the same when one is completely transformed into the other. What I mean to say is, that if there is a rotational energy of E, all of that energy is going to be used to translate the wheel since it can't be lost to friction and the wheel doesn't slip. Then the rotational energy E will go to 0, and the translational energy which was 0, because the wheel wasn't moving before, is now E.
$endgroup$
– Lespiegle
Jan 31 at 18:41
$begingroup$
The same way potential and kinetic energy are the same when one is completely transformed into the other. What I mean to say is, that if there is a rotational energy of E, all of that energy is going to be used to translate the wheel since it can't be lost to friction and the wheel doesn't slip. Then the rotational energy E will go to 0, and the translational energy which was 0, because the wheel wasn't moving before, is now E.
$endgroup$
– Lespiegle
Jan 31 at 18:41
$begingroup$
I understand the dynamics, but you can't just drop the rotational energy from the equation?
$endgroup$
– copper.hat
Jan 31 at 18:57
$begingroup$
I understand the dynamics, but you can't just drop the rotational energy from the equation?
$endgroup$
– copper.hat
Jan 31 at 18:57
$begingroup$
I don't think you do. Let me clarify what I typed earlier. If there is no energy lost to heat then all of the energy created by friction has to be transformed into translational energy or rotational energy. Do you agree with this? Second, the wheel cannot slip. That means that it cannot rotate without creating friction. Therefore it cannot rotate without translating. Do you also agree with this? This is the connection I'm trying to make. I'm not dropping the rotational energy, I'm saying that all of it becomes frictional energy, and then that friction energy translates the wheel.
$endgroup$
– Lespiegle
Jan 31 at 21:15
$begingroup$
I don't think you do. Let me clarify what I typed earlier. If there is no energy lost to heat then all of the energy created by friction has to be transformed into translational energy or rotational energy. Do you agree with this? Second, the wheel cannot slip. That means that it cannot rotate without creating friction. Therefore it cannot rotate without translating. Do you also agree with this? This is the connection I'm trying to make. I'm not dropping the rotational energy, I'm saying that all of it becomes frictional energy, and then that friction energy translates the wheel.
$endgroup$
– Lespiegle
Jan 31 at 21:15
$begingroup$
(I ran out of room in the last comment). Therefore, one can write either the rotational kinetic energy OR the translational kinetic energy and it will account for both of them simultaneously. Writing both of them would likewise be wrong because you would be counting the kinetic energy of the wheel twice.
$endgroup$
– Lespiegle
Jan 31 at 21:17
$begingroup$
(I ran out of room in the last comment). Therefore, one can write either the rotational kinetic energy OR the translational kinetic energy and it will account for both of them simultaneously. Writing both of them would likewise be wrong because you would be counting the kinetic energy of the wheel twice.
$endgroup$
– Lespiegle
Jan 31 at 21:17
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
This is more a physics problem then math. But here are some points that you should consider:
- The moment of inertia of a filled disk is not $J=mr^2$, but $J=frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=frac 12 frac12 mr^2frac{v^2}{r^2}=frac 14 mv^2$$
You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong? - Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=frac12(m/2+m/2)v^2=frac{mv^2}2$$
But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$frac 12 frac m2 0^2+frac 12frac m2 (2v)^2=mv^2ne frac{mv^2}2$$
So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
$endgroup$
$begingroup$
Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
$endgroup$
– Lespiegle
Feb 1 at 13:25
$begingroup$
Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
$endgroup$
– Andrei
Feb 1 at 13:34
add a comment |
$begingroup$
The energy of the system at a configuration $(x,dot{x}, dot{theta})$ is given by
$E(x,dot{x}, dot{theta}) = {1 over 2} m dot{x}^2 + {1 over 2} J dot{theta}^2 + {1 over 2} k x$.
In this particular example, there is an additional constraint which is that
$dot{x} = r dot{theta}$.
When the object rolls, it will 'see' an effective mass of $m+ {J over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J over r^2}$.
The formula for the resonant frequency follows from the usual formula.
Aside:
Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.
The kinetic energies are (i) ${1 over 2} m dot{x}^2$ and (ii) $m dot{x}^2$ (regardless of radius).
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
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$begingroup$
Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
$endgroup$
– Lespiegle
Feb 1 at 13:29
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Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
$endgroup$
– Lespiegle
Feb 1 at 15:51
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Feb 4 at 4:39
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
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active
oldest
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$begingroup$
This is more a physics problem then math. But here are some points that you should consider:
- The moment of inertia of a filled disk is not $J=mr^2$, but $J=frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=frac 12 frac12 mr^2frac{v^2}{r^2}=frac 14 mv^2$$
You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong? - Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=frac12(m/2+m/2)v^2=frac{mv^2}2$$
But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$frac 12 frac m2 0^2+frac 12frac m2 (2v)^2=mv^2ne frac{mv^2}2$$
So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
$endgroup$
$begingroup$
Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
$endgroup$
– Lespiegle
Feb 1 at 13:25
$begingroup$
Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
$endgroup$
– Andrei
Feb 1 at 13:34
add a comment |
$begingroup$
This is more a physics problem then math. But here are some points that you should consider:
- The moment of inertia of a filled disk is not $J=mr^2$, but $J=frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=frac 12 frac12 mr^2frac{v^2}{r^2}=frac 14 mv^2$$
You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong? - Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=frac12(m/2+m/2)v^2=frac{mv^2}2$$
But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$frac 12 frac m2 0^2+frac 12frac m2 (2v)^2=mv^2ne frac{mv^2}2$$
So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
$endgroup$
$begingroup$
Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
$endgroup$
– Lespiegle
Feb 1 at 13:25
$begingroup$
Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
$endgroup$
– Andrei
Feb 1 at 13:34
add a comment |
$begingroup$
This is more a physics problem then math. But here are some points that you should consider:
- The moment of inertia of a filled disk is not $J=mr^2$, but $J=frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=frac 12 frac12 mr^2frac{v^2}{r^2}=frac 14 mv^2$$
You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong? - Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=frac12(m/2+m/2)v^2=frac{mv^2}2$$
But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$frac 12 frac m2 0^2+frac 12frac m2 (2v)^2=mv^2ne frac{mv^2}2$$
So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
$endgroup$
This is more a physics problem then math. But here are some points that you should consider:
- The moment of inertia of a filled disk is not $J=mr^2$, but $J=frac 12 mr^2$, so the kinetic energy of rotation would be half of the kinetic energy of translation. $$K.E.R.=frac 12 frac12 mr^2frac{v^2}{r^2}=frac 14 mv^2$$
You have $J=mr^2$ only if the mass of the wheel is uniformly distributed on the periphery of the disk. So bodies with the same mass can have different moments of inertia. If the kinetic energy of translation is the same as the kinetic energy of rotation, which formula would you use? Why is the other formula wrong? - Let's look at a simpler case, a wheel rolling with constant velocity $v$. The mass of the disk is concentrated in two points, on opposite sides of the disk. We have at each of these points a mass $m/2$. The kinetic energy of the center of the mass is $$K.E._{CM}=frac12(m/2+m/2)v^2=frac{mv^2}2$$
But this must be equal to the sum of kinetic energies of the two particles. When one of the particle is at the bottom, the velocity of that particle is $v_b=0$. At the same time, the velocity of the particle at the top is $v_t=2v$. The sum of kinetic energies is then $$frac 12 frac m2 0^2+frac 12frac m2 (2v)^2=mv^2ne frac{mv^2}2$$
So if the sum is $mv^2$ but the kinetic energy of the center of mass is only half, where did the other half go? The answer is in the kinetic energy of rotation.
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
answered Feb 1 at 5:04
AndreiAndrei
13.5k21230
13.5k21230
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Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
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– Lespiegle
Feb 1 at 13:25
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Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
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– Andrei
Feb 1 at 13:34
add a comment |
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Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
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– Lespiegle
Feb 1 at 13:25
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Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
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– Andrei
Feb 1 at 13:34
$begingroup$
Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
$endgroup$
– Lespiegle
Feb 1 at 13:25
$begingroup$
Thank you for your reply. To clarify, I wasn't saying that either one was wrong. I meant to say that because they describe the same quantity of energy (in this case, not in general) then writing both of them would be the same as writing "2E" when E is the total kinetic energy. But I had forgotten that the value of inertia changes with the distribution of the mass (your first point), and your second point looks exactly like the counter-proof I need to prove to myself that they are in fact separate quantities of energy. I will comment again after I've carefully considered it.
$endgroup$
– Lespiegle
Feb 1 at 13:25
$begingroup$
Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
$endgroup$
– Andrei
Feb 1 at 13:34
$begingroup$
Maybe I should have added a following clarification: We know that the disk can have both translation kinetic energy and rotation kinetic energy at the same time (we can for example suspend the disk from the center of the mass, which we move with speed $v$, then spin it with frequency $omega$, unrelated to $v$). Then the two kinetic energies are unrelated. The fact that the wheel is rolling without slipping just provides a numerical relationship between the two (it will constrain the ratio).
$endgroup$
– Andrei
Feb 1 at 13:34
add a comment |
$begingroup$
The energy of the system at a configuration $(x,dot{x}, dot{theta})$ is given by
$E(x,dot{x}, dot{theta}) = {1 over 2} m dot{x}^2 + {1 over 2} J dot{theta}^2 + {1 over 2} k x$.
In this particular example, there is an additional constraint which is that
$dot{x} = r dot{theta}$.
When the object rolls, it will 'see' an effective mass of $m+ {J over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J over r^2}$.
The formula for the resonant frequency follows from the usual formula.
Aside:
Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.
The kinetic energies are (i) ${1 over 2} m dot{x}^2$ and (ii) $m dot{x}^2$ (regardless of radius).
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
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Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
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– Lespiegle
Feb 1 at 13:29
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Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
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– Lespiegle
Feb 1 at 15:51
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Why the downvote?
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– copper.hat
Feb 4 at 4:39
add a comment |
$begingroup$
The energy of the system at a configuration $(x,dot{x}, dot{theta})$ is given by
$E(x,dot{x}, dot{theta}) = {1 over 2} m dot{x}^2 + {1 over 2} J dot{theta}^2 + {1 over 2} k x$.
In this particular example, there is an additional constraint which is that
$dot{x} = r dot{theta}$.
When the object rolls, it will 'see' an effective mass of $m+ {J over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J over r^2}$.
The formula for the resonant frequency follows from the usual formula.
Aside:
Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.
The kinetic energies are (i) ${1 over 2} m dot{x}^2$ and (ii) $m dot{x}^2$ (regardless of radius).
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
$endgroup$
$begingroup$
Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
$endgroup$
– Lespiegle
Feb 1 at 13:29
$begingroup$
Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
$endgroup$
– Lespiegle
Feb 1 at 15:51
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Feb 4 at 4:39
add a comment |
$begingroup$
The energy of the system at a configuration $(x,dot{x}, dot{theta})$ is given by
$E(x,dot{x}, dot{theta}) = {1 over 2} m dot{x}^2 + {1 over 2} J dot{theta}^2 + {1 over 2} k x$.
In this particular example, there is an additional constraint which is that
$dot{x} = r dot{theta}$.
When the object rolls, it will 'see' an effective mass of $m+ {J over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J over r^2}$.
The formula for the resonant frequency follows from the usual formula.
Aside:
Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.
The kinetic energies are (i) ${1 over 2} m dot{x}^2$ and (ii) $m dot{x}^2$ (regardless of radius).
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
$endgroup$
The energy of the system at a configuration $(x,dot{x}, dot{theta})$ is given by
$E(x,dot{x}, dot{theta}) = {1 over 2} m dot{x}^2 + {1 over 2} J dot{theta}^2 + {1 over 2} k x$.
In this particular example, there is an additional constraint which is that
$dot{x} = r dot{theta}$.
When the object rolls, it will 'see' an effective mass of $m+ {J over r^2}$. That is, it will behave as a single point mass with no inertial component of mass $m+ {J over r^2}$.
The formula for the resonant frequency follows from the usual formula.
Aside:
Consider two systems of mass $m$, (i) is just a frictionless sliding mass and (ii) is an ideal, non slipping hoop with the mass at any positive radius.
The kinetic energies are (i) ${1 over 2} m dot{x}^2$ and (ii) $m dot{x}^2$ (regardless of radius).
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
edited Feb 1 at 0:15
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
answered Jan 31 at 23:55
copper.hatcopper.hat
128k561161
128k561161
$begingroup$
Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
$endgroup$
– Lespiegle
Feb 1 at 13:29
$begingroup$
Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
$endgroup$
– Lespiegle
Feb 1 at 15:51
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Feb 4 at 4:39
add a comment |
$begingroup$
Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
$endgroup$
– Lespiegle
Feb 1 at 13:29
$begingroup$
Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
$endgroup$
– Lespiegle
Feb 1 at 15:51
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Feb 4 at 4:39
$begingroup$
Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
$endgroup$
– Lespiegle
Feb 1 at 13:29
$begingroup$
Thank you for taking the time to illustrate your counter-proof. I will re-consider the problem with yours and Andrei's responses in mind.
$endgroup$
– Lespiegle
Feb 1 at 13:29
$begingroup$
Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
$endgroup$
– Lespiegle
Feb 1 at 15:51
$begingroup$
Yes, I see it now. I tried to say that because of the constraint you gave, that dx/dt is equivalent to dw/dt, then they would necessarily be the same quantity. But when expressed as a mathematical formula E(x,dx/dt, and dw/dt) its obvious that substituting dx/dt for r*dw/dt doesn't eliminate the translational quantity. Andrei's example of two points of mass on a disk was also a great way to conceptualize this. Thank you both very much copper.hat and @Andrei for correcting me. Now I know how to model it with an equation and how to conceptualize J.
$endgroup$
– Lespiegle
Feb 1 at 15:51
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Feb 4 at 4:39
$begingroup$
Why the downvote?
$endgroup$
– copper.hat
Feb 4 at 4:39
add a comment |
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How can the translational and rotational energy be the same quantity???
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– copper.hat
Jan 31 at 16:43
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The same way potential and kinetic energy are the same when one is completely transformed into the other. What I mean to say is, that if there is a rotational energy of E, all of that energy is going to be used to translate the wheel since it can't be lost to friction and the wheel doesn't slip. Then the rotational energy E will go to 0, and the translational energy which was 0, because the wheel wasn't moving before, is now E.
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– Lespiegle
Jan 31 at 18:41
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I understand the dynamics, but you can't just drop the rotational energy from the equation?
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– copper.hat
Jan 31 at 18:57
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I don't think you do. Let me clarify what I typed earlier. If there is no energy lost to heat then all of the energy created by friction has to be transformed into translational energy or rotational energy. Do you agree with this? Second, the wheel cannot slip. That means that it cannot rotate without creating friction. Therefore it cannot rotate without translating. Do you also agree with this? This is the connection I'm trying to make. I'm not dropping the rotational energy, I'm saying that all of it becomes frictional energy, and then that friction energy translates the wheel.
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– Lespiegle
Jan 31 at 21:15
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(I ran out of room in the last comment). Therefore, one can write either the rotational kinetic energy OR the translational kinetic energy and it will account for both of them simultaneously. Writing both of them would likewise be wrong because you would be counting the kinetic energy of the wheel twice.
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– Lespiegle
Jan 31 at 21:17