How does author derives $cos theta = frac{dx}{ds} qquad sintheta = frac{dy }{ds}, $?
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In the book of Dynamics by Horace Lamb, at page 103, is it given
that for a motion on a smooth curve, the equation of motion is given
by $$mv frac{dv }{ ds} = -mg sin theta qquad frac{ mv^2 }{r} =
-mg costheta + R, $$ where $theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by
the curve.
[...]
Then, we have $$cos theta = frac{dx}{ds} qquad sintheta =
frac{dy }{ds}, $$ where $s$ is length of the path taken over the
surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as
upward.
However, I cannot understand how does the author derives the latter equations between $theta$ and the derivatives of $x,y$ wrt $s$.
classical-mechanics
asked Nov 17 at 18:39
onurcanbektas
3,2691935
add a comment |
up vote
2
down vote
favorite
In the book of Dynamics by Horace Lamb, at page 103, is it given
that for a motion on a smooth curve, the equation of motion is given
by $$mv frac{dv }{ ds} = -mg sin theta qquad frac{ mv^2 }{r} =
-mg costheta + R, $$ where $theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by
the curve.
[...]
Then, we have $$cos theta = frac{dx}{ds} qquad sintheta =
frac{dy }{ds}, $$ where $s$ is length of the path taken over the
surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as
upward.
However, I cannot understand how does the author derives the latter equations between $theta$ and the derivatives of $x,y$ wrt $s$.
classical-mechanics
asked Nov 17 at 18:39
onurcanbektas
3,2691935
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the book of Dynamics by Horace Lamb, at page 103, is it given
that for a motion on a smooth curve, the equation of motion is given
by $$mv frac{dv }{ ds} = -mg sin theta qquad frac{ mv^2 }{r} =
-mg costheta + R, $$ where $theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by
the curve.
[...]
Then, we have $$cos theta = frac{dx}{ds} qquad sintheta =
frac{dy }{ds}, $$ where $s$ is length of the path taken over the
surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as
upward.
However, I cannot understand how does the author derives the latter equations between $theta$ and the derivatives of $x,y$ wrt $s$.
classical-mechanics
asked Nov 17 at 18:39
onurcanbektas
3,2691935
In the book of Dynamics by Horace Lamb, at page 103, is it given
that for a motion on a smooth curve, the equation of motion is given
by $$mv frac{dv }{ ds} = -mg sin theta qquad frac{ mv^2 }{r} =
-mg costheta + R, $$ where $theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by
the curve.
[...]
Then, we have $$cos theta = frac{dx}{ds} qquad sintheta =
frac{dy }{ds}, $$ where $s$ is length of the path taken over the
surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as
upward.
However, I cannot understand how does the author derives the latter equations between $theta$ and the derivatives of $x,y$ wrt $s$.
classical-mechanics
classical-mechanics
asked Nov 17 at 18:39
onurcanbektas
3,2691935
asked Nov 17 at 18:39
onurcanbektas
3,2691935
asked Nov 17 at 18:39
onurcanbektas
3,2691935
asked Nov 17 at 18:39
onurcanbektas
3,2691935
asked Nov 17 at 18:39
onurcanbektas
3,2691935
3,2691935
add a comment |
add a comment |
2 Answers
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Hint Write it as $dx=dscostheta$ and $dy=dssintheta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes
answered Nov 17 at 18:47
Andrei
10.3k21025
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
add a comment |
up vote
2
down vote
Now apply the rule of $sin theta$ and $cos theta$
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint Write it as $dx=dscostheta$ and $dy=dssintheta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes
answered Nov 17 at 18:47
Andrei
10.3k21025
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
add a comment |
up vote
1
down vote
accepted
Hint Write it as $dx=dscostheta$ and $dy=dssintheta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes
answered Nov 17 at 18:47
Andrei
10.3k21025
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint Write it as $dx=dscostheta$ and $dy=dssintheta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes
answered Nov 17 at 18:47
Andrei
10.3k21025
Hint Write it as $dx=dscostheta$ and $dy=dssintheta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes
answered Nov 17 at 18:47
Andrei
10.3k21025
answered Nov 17 at 18:47
Andrei
10.3k21025
answered Nov 17 at 18:47
Andrei
10.3k21025
answered Nov 17 at 18:47
Andrei
10.3k21025
10.3k21025
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
add a comment |
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
but the path is not a straight line; even though we can approximate it to be like that.
– onurcanbektas
Nov 17 at 18:51
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
The full path is not straight, but a small element $ds$ can be considered as such.
– Andrei
Nov 17 at 18:53
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Can we assure ourselves that such a consideration will lead us to the correct equation of motion in the global case ? i.e the variation will go to zero, and not to infinity ?
– onurcanbektas
Nov 17 at 18:55
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Yes. In the limit when $dsto 0$, the path can be considered straight. I forgot to add it to my answer, but if the angle between normal and $y$ axis is $theta$, then so is the angle between the tangent to the curve (along $ds$) and $x$ axis.
– Andrei
Nov 17 at 18:57
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
Unless, of course, you allow discontinuity in the derivative of the path. If the path is smooth, then you are OK
– Andrei
Nov 17 at 18:59
add a comment |
up vote
2
down vote
Now apply the rule of $sin theta$ and $cos theta$
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
add a comment |
up vote
2
down vote
Now apply the rule of $sin theta$ and $cos theta$
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
add a comment |
up vote
2
down vote
up vote
2
down vote
Now apply the rule of $sin theta$ and $cos theta$
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
Now apply the rule of $sin theta$ and $cos theta$
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
answered Nov 17 at 18:52
Rakibul Islam Prince
819211
819211
add a comment |
add a comment |
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