Mixing
Equation of motion in a disk and slider system
$begingroup$
I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)
I have used relative velocity principle to calculate velocity of slider A:
$$vec V_C=Rdottheta hat i $$
$$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
$$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
Therfore:
$$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
And as we know the slider has no vertical motion so:
$$Rdotthetacostheta+2.5Rdotphicosphi =0$$
$$dotthetacostheta=-2.5dotphicosphi $$
Therefore:
$$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
From geometry we know:
$$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
$$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
If we want the acceleration in point A:
$$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
So the equation of motion can be derived using newton rule:
$$sum vec F=mvec a $$
$$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$
Is my solution correct?
dynamical-systems
asked Jan 6 at 11:48
H.HH.H
1466
$endgroup$
add a comment |
$begingroup$
I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)
I have used relative velocity principle to calculate velocity of slider A:
$$vec V_C=Rdottheta hat i $$
$$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
$$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
Therfore:
$$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
And as we know the slider has no vertical motion so:
$$Rdotthetacostheta+2.5Rdotphicosphi =0$$
$$dotthetacostheta=-2.5dotphicosphi $$
Therefore:
$$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
From geometry we know:
$$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
$$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
If we want the acceleration in point A:
$$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
So the equation of motion can be derived using newton rule:
$$sum vec F=mvec a $$
$$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$
Is my solution correct?
dynamical-systems
asked Jan 6 at 11:48
H.HH.H
1466
$endgroup$
add a comment |
$begingroup$
I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)
I have used relative velocity principle to calculate velocity of slider A:
$$vec V_C=Rdottheta hat i $$
$$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
$$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
Therfore:
$$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
And as we know the slider has no vertical motion so:
$$Rdotthetacostheta+2.5Rdotphicosphi =0$$
$$dotthetacostheta=-2.5dotphicosphi $$
Therefore:
$$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
From geometry we know:
$$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
$$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
If we want the acceleration in point A:
$$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
So the equation of motion can be derived using newton rule:
$$sum vec F=mvec a $$
$$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$
Is my solution correct?
dynamical-systems
asked Jan 6 at 11:48
H.HH.H
1466
$endgroup$
I want to derive the equation of motion in this system: (the slider mass is m and the disk mass is M and the connecting bar is massless)
I have used relative velocity principle to calculate velocity of slider A:
$$vec V_C=Rdottheta hat i $$
$$vec V_B=vec V_C+vec V_{B/C} =Rdottheta (1+sintheta) hat i +Rdotthetacostheta hat j$$
$$vec V_B=vec V_A+vec V_{B/A}=vec V_A+2.5Rdotphisinphi hat i -2.5Rdotphicosphihat j$$
Therfore:
$$vec V_A=[Rdottheta (1+sintheta)-2.5Rdotphisinphi] hat i +[Rdotthetacostheta+2.5Rdotphicosphi ]hat j$$
And as we know the slider has no vertical motion so:
$$Rdotthetacostheta+2.5Rdotphicosphi =0$$
$$dotthetacostheta=-2.5dotphicosphi $$
Therefore:
$$vec V_A=Rdottheta (1+sintheta+frac {costheta}{cosphi})hat i$$
From geometry we know:
$$Rsintheta =2.5RsinphiRightarrow sintheta =2.5sinphi$$
$$cosphi =sqrt{1-sin^2phi}=sqrt{1-frac {1}{2.5^2}sin^2theta}=1+frac{1}{25}cos2theta $$
If we want the acceleration in point A:
$$vec a_A=frac {d}{dt}vec V_A=[Rddottheta (1+sintheta+costheta)+Rdottheta^2 (costheta-sintheta)]hat i$$
So the equation of motion can be derived using newton rule:
$$sum vec F=mvec a $$
$$F (t)= mRddottheta (1+sintheta+costheta)+mRdottheta^2 (costheta-sintheta)$$
Is my solution correct?
dynamical-systems
dynamical-systems
asked Jan 6 at 11:48
H.HH.H
1466
asked Jan 6 at 11:48
H.HH.H
1466
edited Jan 6 at 18:29
H.H
asked Jan 6 at 11:48
H.HH.H
1466
asked Jan 6 at 11:48
H.HH.H
1466
asked Jan 6 at 11:48
H.HH.H
1466
1466
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
$$V=0$$
$$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
$$I=frac{3}{2}MR^2$$
if your answer for velocity of the slider is correct we can write:
$$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$
so if you use Lagrange equations, you can find the answer:
$$L=T-V$$
$${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$
answered Jan 6 at 21:48
HarryHarry
283
$endgroup$
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
add a comment |
$begingroup$
The total kinetic energy is given by
$$
K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
$$
with $omega = dottheta$
We know also that
$$
vec v_B = vec v_C + vecomegatimes(B-C)\
vec v_B = vec v_A + vec {dotphi} times (B-A)
$$
or
$$
vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
$$
Here
$$
B-C = R(costheta,sintheta)\
B-A = lambda R(-cosphi,sinphi)\
vec v_C = R(omega,0)\
vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
$$
then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as
$$
T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
$$
Here $eta = (theta,phi)$
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
$$V=0$$
$$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
$$I=frac{3}{2}MR^2$$
if your answer for velocity of the slider is correct we can write:
$$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$
so if you use Lagrange equations, you can find the answer:
$$L=T-V$$
$${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$
answered Jan 6 at 21:48
HarryHarry
283
$endgroup$
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
add a comment |
$begingroup$
It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
$$V=0$$
$$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
$$I=frac{3}{2}MR^2$$
if your answer for velocity of the slider is correct we can write:
$$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$
so if you use Lagrange equations, you can find the answer:
$$L=T-V$$
$${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$
answered Jan 6 at 21:48
HarryHarry
283
$endgroup$
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
add a comment |
$begingroup$
It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
$$V=0$$
$$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
$$I=frac{3}{2}MR^2$$
if your answer for velocity of the slider is correct we can write:
$$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$
so if you use Lagrange equations, you can find the answer:
$$L=T-V$$
$${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$
answered Jan 6 at 21:48
HarryHarry
283
$endgroup$
It seems like a better idea if you try Lagrange method by deriving kinetic and potential energies:
$$V=0$$
$$T=frac{1}{2}mV_A^2+frac{1}{2}I_{disk}omega^2$$
$$I=frac{3}{2}MR^2$$
if your answer for velocity of the slider is correct we can write:
$$T=frac{1}{2}m[Rdottheta (1+sintheta+frac {costheta}{cosphi})]^2+frac{1}{2}frac{3}{2}MR^2dottheta^2$$
so if you use Lagrange equations, you can find the answer:
$$L=T-V$$
$${displaystyle {frac {mathrm {d} }{mathrm {d} t}}left({frac {partial L}{partial {dot {q}}_{j}}}right)={frac {partial L}{partial q_{j}}}}$$
answered Jan 6 at 21:48
HarryHarry
283
answered Jan 6 at 21:48
HarryHarry
283
answered Jan 6 at 21:48
HarryHarry
283
answered Jan 6 at 21:48
HarryHarry
283
283
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
add a comment |
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
$begingroup$
Thank you for your suggestion, but can you derive the equation of motion in order to compare two answers?
$endgroup$
– H.H
Jan 6 at 21:57
add a comment |
$begingroup$
The total kinetic energy is given by
$$
K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
$$
with $omega = dottheta$
We know also that
$$
vec v_B = vec v_C + vecomegatimes(B-C)\
vec v_B = vec v_A + vec {dotphi} times (B-A)
$$
or
$$
vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
$$
Here
$$
B-C = R(costheta,sintheta)\
B-A = lambda R(-cosphi,sinphi)\
vec v_C = R(omega,0)\
vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
$$
then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as
$$
T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
$$
Here $eta = (theta,phi)$
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
$endgroup$
add a comment |
$begingroup$
The total kinetic energy is given by
$$
K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
$$
with $omega = dottheta$
We know also that
$$
vec v_B = vec v_C + vecomegatimes(B-C)\
vec v_B = vec v_A + vec {dotphi} times (B-A)
$$
or
$$
vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
$$
Here
$$
B-C = R(costheta,sintheta)\
B-A = lambda R(-cosphi,sinphi)\
vec v_C = R(omega,0)\
vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
$$
then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as
$$
T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
$$
Here $eta = (theta,phi)$
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
$endgroup$
add a comment |
$begingroup$
The total kinetic energy is given by
$$
K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
$$
with $omega = dottheta$
We know also that
$$
vec v_B = vec v_C + vecomegatimes(B-C)\
vec v_B = vec v_A + vec {dotphi} times (B-A)
$$
or
$$
vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
$$
Here
$$
B-C = R(costheta,sintheta)\
B-A = lambda R(-cosphi,sinphi)\
vec v_C = R(omega,0)\
vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
$$
then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as
$$
T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
$$
Here $eta = (theta,phi)$
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
$endgroup$
The total kinetic energy is given by
$$
K = frac 12 M||vec v_C||^2+frac 12 J_C omega^2+frac 12 m ||vec v_A||^2
$$
with $omega = dottheta$
We know also that
$$
vec v_B = vec v_C + vecomegatimes(B-C)\
vec v_B = vec v_A + vec {dotphi} times (B-A)
$$
or
$$
vec v_C + vecomegatimes(B-C) = vec v_A + vec {dotphi} times (B-A)
$$
Here
$$
B-C = R(costheta,sintheta)\
B-A = lambda R(-cosphi,sinphi)\
vec v_C = R(omega,0)\
vec v_A = R (omega-(lambdadotphi+omega)sintheta,(omega-lambdadotphi)costheta)
$$
then plugin those results into the kinetic energy expression and considering that $T = K - V$ with $V = 0$ we can derive the movement equations as
$$
T_{eta}-frac{d}{dt}T_{doteta}=mathcal{F}
$$
Here $eta = (theta,phi)$
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
edited Jan 10 at 18:22
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
answered Jan 10 at 14:06
CesareoCesareo
8,6733516
8,6733516
add a comment |
add a comment |
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