Why are the units of angular acceleration the same as that of angular velocity squared?
$begingroup$
According to this answer, the units for angular velocity squared are $mathrm{rad}/s^2$. The units for angular acceleration are also $mathrm{rad}/s^2$. Why is this the case?
acceleration units angular-velocity
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
$endgroup$
add a comment |
$begingroup$
According to this answer, the units for angular velocity squared are $mathrm{rad}/s^2$. The units for angular acceleration are also $mathrm{rad}/s^2$. Why is this the case?
acceleration units angular-velocity
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
$endgroup$
4
$begingroup$
Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrm{rad}/s^2$ (they would be $(mathrm{rad}/s)^2$).
$endgroup$
– Chemomechanics
Dec 5 '18 at 15:55
$begingroup$
Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
$endgroup$
– knzhou
Dec 5 '18 at 16:44
$begingroup$
Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
add a comment |
$begingroup$
According to this answer, the units for angular velocity squared are $mathrm{rad}/s^2$. The units for angular acceleration are also $mathrm{rad}/s^2$. Why is this the case?
acceleration units angular-velocity
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
$endgroup$
According to this answer, the units for angular velocity squared are $mathrm{rad}/s^2$. The units for angular acceleration are also $mathrm{rad}/s^2$. Why is this the case?
acceleration units angular-velocity
acceleration units angular-velocity
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
edited Dec 5 '18 at 20:57
Ruslan
9,27643070
9,27643070
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
asked Dec 5 '18 at 15:42
Raymo111Raymo111
1417
1417
4
$begingroup$
Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrm{rad}/s^2$ (they would be $(mathrm{rad}/s)^2$).
$endgroup$
– Chemomechanics
Dec 5 '18 at 15:55
$begingroup$
Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
$endgroup$
– knzhou
Dec 5 '18 at 16:44
$begingroup$
Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
add a comment |
4
$begingroup$
Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrm{rad}/s^2$ (they would be $(mathrm{rad}/s)^2$).
$endgroup$
– Chemomechanics
Dec 5 '18 at 15:55
$begingroup$
Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
$endgroup$
– knzhou
Dec 5 '18 at 16:44
$begingroup$
Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
4
4
$begingroup$
Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrm{rad}/s^2$ (they would be $(mathrm{rad}/s)^2$).
$endgroup$
– Chemomechanics
Dec 5 '18 at 15:55
$begingroup$
Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrm{rad}/s^2$ (they would be $(mathrm{rad}/s)^2$).
$endgroup$
– Chemomechanics
Dec 5 '18 at 15:55
$begingroup$
Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
$endgroup$
– knzhou
Dec 5 '18 at 16:44
$begingroup$
Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
$endgroup$
– knzhou
Dec 5 '18 at 16:44
$begingroup$
Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
$endgroup$
– Cort Ammon
Dec 5 '18 at 20:12
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
They have the same units of $mathrm s^{-2}$ only if you don't use $mathrm{rad}$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrm{rad}/mathrm s^2$, and that of square of angular velocity is $mathrm{rad}^2/mathrm s^2$.
If radian is dimensionless, why do we not "simplify" $mathrm{rad}^2$ to $mathrm{rad}$? For the same reason as why we introduced $mathrm{rad}$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrm{rad}^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
$endgroup$
add a comment |
$begingroup$
By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.
Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
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add a comment |
$begingroup$
They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^{-2}$ and when you derive you get the same.
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
$endgroup$
4
$begingroup$
Yes, but my question was aboutwhythey were the same, notifthey were the same.
$endgroup$
– Raymo111
Dec 5 '18 at 15:49
add a comment |
$begingroup$
Radians are dimensionless. You can safely set $text{rad}to 1$. Using $[x]$ to mean "the units of $x$, we find:
The units of angular velocity $[omega] = left[frac{dtheta}{dt}right]= frac{1}{T}$ where $T$ is time. Then $[omega^2] = frac{1}{T^2}$.
The units of angular acceleration $[alpha] = left[frac{d^2theta}{dt^2}right] = frac{1}{T^2}$.
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
$endgroup$
add a comment |
$begingroup$
Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).
Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.
We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
They have the same units of $mathrm s^{-2}$ only if you don't use $mathrm{rad}$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrm{rad}/mathrm s^2$, and that of square of angular velocity is $mathrm{rad}^2/mathrm s^2$.
If radian is dimensionless, why do we not "simplify" $mathrm{rad}^2$ to $mathrm{rad}$? For the same reason as why we introduced $mathrm{rad}$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrm{rad}^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
$endgroup$
add a comment |
$begingroup$
They have the same units of $mathrm s^{-2}$ only if you don't use $mathrm{rad}$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrm{rad}/mathrm s^2$, and that of square of angular velocity is $mathrm{rad}^2/mathrm s^2$.
If radian is dimensionless, why do we not "simplify" $mathrm{rad}^2$ to $mathrm{rad}$? For the same reason as why we introduced $mathrm{rad}$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrm{rad}^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
$endgroup$
add a comment |
$begingroup$
They have the same units of $mathrm s^{-2}$ only if you don't use $mathrm{rad}$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrm{rad}/mathrm s^2$, and that of square of angular velocity is $mathrm{rad}^2/mathrm s^2$.
If radian is dimensionless, why do we not "simplify" $mathrm{rad}^2$ to $mathrm{rad}$? For the same reason as why we introduced $mathrm{rad}$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrm{rad}^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
$endgroup$
They have the same units of $mathrm s^{-2}$ only if you don't use $mathrm{rad}$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrm{rad}/mathrm s^2$, and that of square of angular velocity is $mathrm{rad}^2/mathrm s^2$.
If radian is dimensionless, why do we not "simplify" $mathrm{rad}^2$ to $mathrm{rad}$? For the same reason as why we introduced $mathrm{rad}$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrm{rad}^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
edited Dec 5 '18 at 20:55
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
answered Dec 5 '18 at 16:10
RuslanRuslan
9,27643070
9,27643070
add a comment |
add a comment |
$begingroup$
By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.
Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
$endgroup$
add a comment |
$begingroup$
By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.
Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
$endgroup$
add a comment |
$begingroup$
By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.
Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
$endgroup$
By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.
Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
answered Dec 5 '18 at 16:22
David WhiteDavid White
4,3282619
4,3282619
add a comment |
add a comment |
$begingroup$
They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^{-2}$ and when you derive you get the same.
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
$endgroup$
4
$begingroup$
Yes, but my question was aboutwhythey were the same, notifthey were the same.
$endgroup$
– Raymo111
Dec 5 '18 at 15:49
add a comment |
$begingroup$
They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^{-2}$ and when you derive you get the same.
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
$endgroup$
4
$begingroup$
Yes, but my question was aboutwhythey were the same, notifthey were the same.
$endgroup$
– Raymo111
Dec 5 '18 at 15:49
add a comment |
$begingroup$
They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^{-2}$ and when you derive you get the same.
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
$endgroup$
They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^{-2}$ and when you derive you get the same.
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
edited Dec 5 '18 at 16:04
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
answered Dec 5 '18 at 15:48
Run like hellRun like hell
1,318725
1,318725
4
$begingroup$
Yes, but my question was aboutwhythey were the same, notifthey were the same.
$endgroup$
– Raymo111
Dec 5 '18 at 15:49
add a comment |
4
$begingroup$
Yes, but my question was aboutwhythey were the same, notifthey were the same.
$endgroup$
– Raymo111
Dec 5 '18 at 15:49
4
4
$begingroup$
Yes, but my question was about
why they were the same, not if they were the same.$endgroup$
– Raymo111
Dec 5 '18 at 15:49
$begingroup$
Yes, but my question was about
why they were the same, not if they were the same.$endgroup$
– Raymo111
Dec 5 '18 at 15:49
add a comment |
$begingroup$
Radians are dimensionless. You can safely set $text{rad}to 1$. Using $[x]$ to mean "the units of $x$, we find:
The units of angular velocity $[omega] = left[frac{dtheta}{dt}right]= frac{1}{T}$ where $T$ is time. Then $[omega^2] = frac{1}{T^2}$.
The units of angular acceleration $[alpha] = left[frac{d^2theta}{dt^2}right] = frac{1}{T^2}$.
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
$endgroup$
add a comment |
$begingroup$
Radians are dimensionless. You can safely set $text{rad}to 1$. Using $[x]$ to mean "the units of $x$, we find:
The units of angular velocity $[omega] = left[frac{dtheta}{dt}right]= frac{1}{T}$ where $T$ is time. Then $[omega^2] = frac{1}{T^2}$.
The units of angular acceleration $[alpha] = left[frac{d^2theta}{dt^2}right] = frac{1}{T^2}$.
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
$endgroup$
add a comment |
$begingroup$
Radians are dimensionless. You can safely set $text{rad}to 1$. Using $[x]$ to mean "the units of $x$, we find:
The units of angular velocity $[omega] = left[frac{dtheta}{dt}right]= frac{1}{T}$ where $T$ is time. Then $[omega^2] = frac{1}{T^2}$.
The units of angular acceleration $[alpha] = left[frac{d^2theta}{dt^2}right] = frac{1}{T^2}$.
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
$endgroup$
Radians are dimensionless. You can safely set $text{rad}to 1$. Using $[x]$ to mean "the units of $x$, we find:
The units of angular velocity $[omega] = left[frac{dtheta}{dt}right]= frac{1}{T}$ where $T$ is time. Then $[omega^2] = frac{1}{T^2}$.
The units of angular acceleration $[alpha] = left[frac{d^2theta}{dt^2}right] = frac{1}{T^2}$.
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
answered Dec 6 '18 at 3:39
zahbazzahbaz
382317
382317
add a comment |
add a comment |
$begingroup$
Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).
Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.
We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
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add a comment |
$begingroup$
Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).
Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.
We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
$endgroup$
add a comment |
$begingroup$
Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).
Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.
We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
$endgroup$
Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).
Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.
We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
answered Dec 6 '18 at 12:36
geoff22873geoff22873
1
1
add a comment |
add a comment |
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Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrm{rad}/s^2$ (they would be $(mathrm{rad}/s)^2$).
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– Chemomechanics
Dec 5 '18 at 15:55
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Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
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– knzhou
Dec 5 '18 at 16:44
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Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
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– Cort Ammon
Dec 5 '18 at 20:12
$begingroup$
In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
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– Cort Ammon
Dec 5 '18 at 20:12