What is the least eccentricity of an ellipse that can rest on a plane inclined at an angle $alpha$ with the...
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A perfectly rough plane is inclined at an angle $alpha$ to the horizontal. What is the least eccentricity of an elliptic cylinder that resists rolling down the plane?
geometry coordinate-systems classical-mechanics
edited Jan 13 at 11:20
Narasimham
21k62158
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
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closed as unclear what you're asking by Narasimham, John Douma, Shailesh, heropup, Cesareo Jan 13 at 9:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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A perfectly rough plane is inclined at an angle $alpha$ to the horizontal. What is the least eccentricity of an elliptic cylinder that resists rolling down the plane?
geometry coordinate-systems classical-mechanics
edited Jan 13 at 11:20
Narasimham
21k62158
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
$endgroup$
closed as unclear what you're asking by Narasimham, John Douma, Shailesh, heropup, Cesareo Jan 13 at 9:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
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I do find this clear. "Perfetly rough" means you can't slide anything on it but you might be able to roll something. The problem is to find conditions where rolling alone causes the center of gravity to have local minima.
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– Oscar Lanzi
Jan 13 at 11:20
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@NagaArjun Please feel free to revert if my edit is not ok.
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– Narasimham
Jan 13 at 11:21
add a comment |
$begingroup$
A perfectly rough plane is inclined at an angle $alpha$ to the horizontal. What is the least eccentricity of an elliptic cylinder that resists rolling down the plane?
geometry coordinate-systems classical-mechanics
edited Jan 13 at 11:20
Narasimham
21k62158
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
$endgroup$
A perfectly rough plane is inclined at an angle $alpha$ to the horizontal. What is the least eccentricity of an elliptic cylinder that resists rolling down the plane?
geometry coordinate-systems classical-mechanics
geometry coordinate-systems classical-mechanics
edited Jan 13 at 11:20
Narasimham
21k62158
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
edited Jan 13 at 11:20
Narasimham
21k62158
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
edited Jan 13 at 11:20
Narasimham
21k62158
edited Jan 13 at 11:20
Narasimham
21k62158
edited Jan 13 at 11:20
Narasimham
21k62158
21k62158
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
asked Jan 13 at 7:36
NagaArjunNagaArjun
162
162
closed as unclear what you're asking by Narasimham, John Douma, Shailesh, heropup, Cesareo Jan 13 at 9:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Narasimham, John Douma, Shailesh, heropup, Cesareo Jan 13 at 9:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I do find this clear. "Perfetly rough" means you can't slide anything on it but you might be able to roll something. The problem is to find conditions where rolling alone causes the center of gravity to have local minima.
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– Oscar Lanzi
Jan 13 at 11:20
$begingroup$
@NagaArjun Please feel free to revert if my edit is not ok.
$endgroup$
– Narasimham
Jan 13 at 11:21
add a comment |
1
$begingroup$
I do find this clear. "Perfetly rough" means you can't slide anything on it but you might be able to roll something. The problem is to find conditions where rolling alone causes the center of gravity to have local minima.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:20
$begingroup$
@NagaArjun Please feel free to revert if my edit is not ok.
$endgroup$
– Narasimham
Jan 13 at 11:21
1
1
$begingroup$
I do find this clear. "Perfetly rough" means you can't slide anything on it but you might be able to roll something. The problem is to find conditions where rolling alone causes the center of gravity to have local minima.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:20
$begingroup$
I do find this clear. "Perfetly rough" means you can't slide anything on it but you might be able to roll something. The problem is to find conditions where rolling alone causes the center of gravity to have local minima.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:20
$begingroup$
@NagaArjun Please feel free to revert if my edit is not ok.
$endgroup$
– Narasimham
Jan 13 at 11:21
$begingroup$
@NagaArjun Please feel free to revert if my edit is not ok.
$endgroup$
– Narasimham
Jan 13 at 11:21
add a comment |
1 Answer
1
active
oldest
votes
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At equilibrium, the weight $W$, the normal reaction $R$ and the friction $F$ should be concurrent (at the point of contact). The inclination $alpha$ is the angle between the radius vector and the corresponding normal.
Unit outward normal vector:
$$mathbf{n}=
frac{(bcos t,asin t)}{sqrt{a^2sin^2 t+b^2cos^2 t}}$$
Unit radius vector:
$$hat{mathbf{r}}=
frac{(acos t,bsin t)}{sqrt{a^2cos^2 t+b^2sin^2 t}}$$
Now
begin{align}
cos alpha &= mathbf{n} cdot hat{mathbf{r}} \
&= frac{ab}{sqrt{(a^2 sin^2 t+b^2 cos^2 t)(a^2cos^2 t+b^2sin^2 t)}} \
&= absqrt{frac{8}{a^4+6a^2b^2+b^4-(a^2-b^2)^2 cos 4t}}
end{align}
The upper and lower bounds are given by $t=npi/2$ and $t=(2n+1)pi/4$ respectively.
That is,
$$frac{2ab}{a^2+b^2} le cos alpha le 1$$
$$cos^{-1} frac{2ab}{a^2+b^2} ge alpha ge 0$$
Equivalently
$$frac{2sqrt{1-e^2}}{2-e^2} le cos alpha$$
On solving,
$$e^2 ge frac{2sin alpha}{1+sin alpha}$$
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
At equilibrium, the weight $W$, the normal reaction $R$ and the friction $F$ should be concurrent (at the point of contact). The inclination $alpha$ is the angle between the radius vector and the corresponding normal.
Unit outward normal vector:
$$mathbf{n}=
frac{(bcos t,asin t)}{sqrt{a^2sin^2 t+b^2cos^2 t}}$$
Unit radius vector:
$$hat{mathbf{r}}=
frac{(acos t,bsin t)}{sqrt{a^2cos^2 t+b^2sin^2 t}}$$
Now
begin{align}
cos alpha &= mathbf{n} cdot hat{mathbf{r}} \
&= frac{ab}{sqrt{(a^2 sin^2 t+b^2 cos^2 t)(a^2cos^2 t+b^2sin^2 t)}} \
&= absqrt{frac{8}{a^4+6a^2b^2+b^4-(a^2-b^2)^2 cos 4t}}
end{align}
The upper and lower bounds are given by $t=npi/2$ and $t=(2n+1)pi/4$ respectively.
That is,
$$frac{2ab}{a^2+b^2} le cos alpha le 1$$
$$cos^{-1} frac{2ab}{a^2+b^2} ge alpha ge 0$$
Equivalently
$$frac{2sqrt{1-e^2}}{2-e^2} le cos alpha$$
On solving,
$$e^2 ge frac{2sin alpha}{1+sin alpha}$$
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
add a comment |
$begingroup$
At equilibrium, the weight $W$, the normal reaction $R$ and the friction $F$ should be concurrent (at the point of contact). The inclination $alpha$ is the angle between the radius vector and the corresponding normal.
Unit outward normal vector:
$$mathbf{n}=
frac{(bcos t,asin t)}{sqrt{a^2sin^2 t+b^2cos^2 t}}$$
Unit radius vector:
$$hat{mathbf{r}}=
frac{(acos t,bsin t)}{sqrt{a^2cos^2 t+b^2sin^2 t}}$$
Now
begin{align}
cos alpha &= mathbf{n} cdot hat{mathbf{r}} \
&= frac{ab}{sqrt{(a^2 sin^2 t+b^2 cos^2 t)(a^2cos^2 t+b^2sin^2 t)}} \
&= absqrt{frac{8}{a^4+6a^2b^2+b^4-(a^2-b^2)^2 cos 4t}}
end{align}
The upper and lower bounds are given by $t=npi/2$ and $t=(2n+1)pi/4$ respectively.
That is,
$$frac{2ab}{a^2+b^2} le cos alpha le 1$$
$$cos^{-1} frac{2ab}{a^2+b^2} ge alpha ge 0$$
Equivalently
$$frac{2sqrt{1-e^2}}{2-e^2} le cos alpha$$
On solving,
$$e^2 ge frac{2sin alpha}{1+sin alpha}$$
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
add a comment |
$begingroup$
At equilibrium, the weight $W$, the normal reaction $R$ and the friction $F$ should be concurrent (at the point of contact). The inclination $alpha$ is the angle between the radius vector and the corresponding normal.
Unit outward normal vector:
$$mathbf{n}=
frac{(bcos t,asin t)}{sqrt{a^2sin^2 t+b^2cos^2 t}}$$
Unit radius vector:
$$hat{mathbf{r}}=
frac{(acos t,bsin t)}{sqrt{a^2cos^2 t+b^2sin^2 t}}$$
Now
begin{align}
cos alpha &= mathbf{n} cdot hat{mathbf{r}} \
&= frac{ab}{sqrt{(a^2 sin^2 t+b^2 cos^2 t)(a^2cos^2 t+b^2sin^2 t)}} \
&= absqrt{frac{8}{a^4+6a^2b^2+b^4-(a^2-b^2)^2 cos 4t}}
end{align}
The upper and lower bounds are given by $t=npi/2$ and $t=(2n+1)pi/4$ respectively.
That is,
$$frac{2ab}{a^2+b^2} le cos alpha le 1$$
$$cos^{-1} frac{2ab}{a^2+b^2} ge alpha ge 0$$
Equivalently
$$frac{2sqrt{1-e^2}}{2-e^2} le cos alpha$$
On solving,
$$e^2 ge frac{2sin alpha}{1+sin alpha}$$
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
$endgroup$
At equilibrium, the weight $W$, the normal reaction $R$ and the friction $F$ should be concurrent (at the point of contact). The inclination $alpha$ is the angle between the radius vector and the corresponding normal.
Unit outward normal vector:
$$mathbf{n}=
frac{(bcos t,asin t)}{sqrt{a^2sin^2 t+b^2cos^2 t}}$$
Unit radius vector:
$$hat{mathbf{r}}=
frac{(acos t,bsin t)}{sqrt{a^2cos^2 t+b^2sin^2 t}}$$
Now
begin{align}
cos alpha &= mathbf{n} cdot hat{mathbf{r}} \
&= frac{ab}{sqrt{(a^2 sin^2 t+b^2 cos^2 t)(a^2cos^2 t+b^2sin^2 t)}} \
&= absqrt{frac{8}{a^4+6a^2b^2+b^4-(a^2-b^2)^2 cos 4t}}
end{align}
The upper and lower bounds are given by $t=npi/2$ and $t=(2n+1)pi/4$ respectively.
That is,
$$frac{2ab}{a^2+b^2} le cos alpha le 1$$
$$cos^{-1} frac{2ab}{a^2+b^2} ge alpha ge 0$$
Equivalently
$$frac{2sqrt{1-e^2}}{2-e^2} le cos alpha$$
On solving,
$$e^2 ge frac{2sin alpha}{1+sin alpha}$$
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
edited Jan 13 at 14:20
Oscar Lanzi
13.2k12136
13.2k12136
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
answered Jan 13 at 8:25
Ng Chung TakNg Chung Tak
14.8k31334
14.8k31334
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
add a comment |
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
Not being stopped but have a position enabling equilibrium. That's not a stable equilibrium indeed.
$endgroup$
– Ng Chung Tak
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
I would next apply the Pythagorean identity to render $sinalphale(e^2)/(2-e^2)$, which lends itself to an explicit expression for $e$ which is what the problem seems to ask.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:15
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NgChungTak can we do it without vector method?
$endgroup$
– NagaArjun
Jan 13 at 11:25
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
@NagaArjun, you may use the equation of normal namely, $$axsec t-bycsc t=a^2-b^2$$
$endgroup$
– Ng Chung Tak
Jan 13 at 11:27
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
$begingroup$
Re-expressed the bounds more intuitively. The mathematics remains the same. Also to prove the bounds, render the squared trig functions in terms of $cos 2t$ and apply $|cos 2t|le 1$.
$endgroup$
– Oscar Lanzi
Jan 13 at 14:22
add a comment |
1
$begingroup$
I do find this clear. "Perfetly rough" means you can't slide anything on it but you might be able to roll something. The problem is to find conditions where rolling alone causes the center of gravity to have local minima.
$endgroup$
– Oscar Lanzi
Jan 13 at 11:20
$begingroup$
@NagaArjun Please feel free to revert if my edit is not ok.
$endgroup$
– Narasimham
Jan 13 at 11:21