Oil Drill's Equation of Motion
$begingroup$
I need to find equation of motion for the following system:
https://i.stack.imgur.com/d236l.jpg
Parameter of the system are $J_T, J_B, theta_T, theta_B, T_f $
I derived two equation, but i am not sure that is true.
For Table/Top:
begin{equation}label{eq:1}
dot{omega}_T(t)=-frac{b}{J_T}omega_T(t)-bigg[frac{k}{J_T}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]+frac{1}{J_T}T(t)
end{equation}
For Bit/Bottom:
begin{equation}label{eq:2}
dot{omega}_B(t)=-frac{b}{J_B}omega_B(t)+bigg[frac{k}{J_B}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]-frac{1}{J_B}T_f(t)
end{equation}
ordinary-differential-equations
asked Dec 15 '18 at 6:20
SquanchSquanch
353
$endgroup$
add a comment |
$begingroup$
I need to find equation of motion for the following system:
https://i.stack.imgur.com/d236l.jpg
Parameter of the system are $J_T, J_B, theta_T, theta_B, T_f $
I derived two equation, but i am not sure that is true.
For Table/Top:
begin{equation}label{eq:1}
dot{omega}_T(t)=-frac{b}{J_T}omega_T(t)-bigg[frac{k}{J_T}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]+frac{1}{J_T}T(t)
end{equation}
For Bit/Bottom:
begin{equation}label{eq:2}
dot{omega}_B(t)=-frac{b}{J_B}omega_B(t)+bigg[frac{k}{J_B}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]-frac{1}{J_B}T_f(t)
end{equation}
ordinary-differential-equations
asked Dec 15 '18 at 6:20
SquanchSquanch
353
$endgroup$
add a comment |
$begingroup$
I need to find equation of motion for the following system:
https://i.stack.imgur.com/d236l.jpg
Parameter of the system are $J_T, J_B, theta_T, theta_B, T_f $
I derived two equation, but i am not sure that is true.
For Table/Top:
begin{equation}label{eq:1}
dot{omega}_T(t)=-frac{b}{J_T}omega_T(t)-bigg[frac{k}{J_T}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]+frac{1}{J_T}T(t)
end{equation}
For Bit/Bottom:
begin{equation}label{eq:2}
dot{omega}_B(t)=-frac{b}{J_B}omega_B(t)+bigg[frac{k}{J_B}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]-frac{1}{J_B}T_f(t)
end{equation}
ordinary-differential-equations
asked Dec 15 '18 at 6:20
SquanchSquanch
353
$endgroup$
I need to find equation of motion for the following system:
https://i.stack.imgur.com/d236l.jpg
Parameter of the system are $J_T, J_B, theta_T, theta_B, T_f $
I derived two equation, but i am not sure that is true.
For Table/Top:
begin{equation}label{eq:1}
dot{omega}_T(t)=-frac{b}{J_T}omega_T(t)-bigg[frac{k}{J_T}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]+frac{1}{J_T}T(t)
end{equation}
For Bit/Bottom:
begin{equation}label{eq:2}
dot{omega}_B(t)=-frac{b}{J_B}omega_B(t)+bigg[frac{k}{J_B}theta_T(t)-frac{k}{J_T}theta_B(t)bigg]-frac{1}{J_B}T_f(t)
end{equation}
ordinary-differential-equations
ordinary-differential-equations
asked Dec 15 '18 at 6:20
SquanchSquanch
353
asked Dec 15 '18 at 6:20
SquanchSquanch
353
asked Dec 15 '18 at 6:20
SquanchSquanch
353
asked Dec 15 '18 at 6:20
SquanchSquanch
353
asked Dec 15 '18 at 6:20
SquanchSquanch
353
353
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Considering the kinetic energy
$$
E = frac 12 J_Bomega_B^2+frac 12 J_Tomega_b^T+frac 12 k(theta_B-theta_T)^2
$$
the motion equations are derived according Euler-Lagrange
$$
frac{partial E}{partial theta}-frac{d}{dt}frac{partial E}{partialomega} = tau
$$
so we have
$$
k(theta_B-theta_T)+J_Bdotomega_B + bomega_B = -T_f\
-k(theta_B-theta_T)+J_Tdotomega_T + bomega_T = T
$$
Here $dotomega = ddottheta$
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
$endgroup$
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Considering the kinetic energy
$$
E = frac 12 J_Bomega_B^2+frac 12 J_Tomega_b^T+frac 12 k(theta_B-theta_T)^2
$$
the motion equations are derived according Euler-Lagrange
$$
frac{partial E}{partial theta}-frac{d}{dt}frac{partial E}{partialomega} = tau
$$
so we have
$$
k(theta_B-theta_T)+J_Bdotomega_B + bomega_B = -T_f\
-k(theta_B-theta_T)+J_Tdotomega_T + bomega_T = T
$$
Here $dotomega = ddottheta$
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
$endgroup$
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
add a comment |
$begingroup$
Considering the kinetic energy
$$
E = frac 12 J_Bomega_B^2+frac 12 J_Tomega_b^T+frac 12 k(theta_B-theta_T)^2
$$
the motion equations are derived according Euler-Lagrange
$$
frac{partial E}{partial theta}-frac{d}{dt}frac{partial E}{partialomega} = tau
$$
so we have
$$
k(theta_B-theta_T)+J_Bdotomega_B + bomega_B = -T_f\
-k(theta_B-theta_T)+J_Tdotomega_T + bomega_T = T
$$
Here $dotomega = ddottheta$
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
$endgroup$
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
add a comment |
$begingroup$
Considering the kinetic energy
$$
E = frac 12 J_Bomega_B^2+frac 12 J_Tomega_b^T+frac 12 k(theta_B-theta_T)^2
$$
the motion equations are derived according Euler-Lagrange
$$
frac{partial E}{partial theta}-frac{d}{dt}frac{partial E}{partialomega} = tau
$$
so we have
$$
k(theta_B-theta_T)+J_Bdotomega_B + bomega_B = -T_f\
-k(theta_B-theta_T)+J_Tdotomega_T + bomega_T = T
$$
Here $dotomega = ddottheta$
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
$endgroup$
Considering the kinetic energy
$$
E = frac 12 J_Bomega_B^2+frac 12 J_Tomega_b^T+frac 12 k(theta_B-theta_T)^2
$$
the motion equations are derived according Euler-Lagrange
$$
frac{partial E}{partial theta}-frac{d}{dt}frac{partial E}{partialomega} = tau
$$
so we have
$$
k(theta_B-theta_T)+J_Bdotomega_B + bomega_B = -T_f\
-k(theta_B-theta_T)+J_Tdotomega_T + bomega_T = T
$$
Here $dotomega = ddottheta$
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
answered Dec 15 '18 at 10:18
CesareoCesareo
8,8293516
8,8293516
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
add a comment |
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
$begingroup$
This verifies my result. Thank you so much.
$endgroup$
– Squanch
Dec 15 '18 at 10:32
add a comment |
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