- #1

mfig

- 282

- 97

Hello,

I found this problem while preparing for the qualifying exams and thought I would tackle it. I have no idea if it is correct, but I would like someone to tell me if there is an easier way to solve this than I used. Perhaps some Lagrangian or Hamiltonian approach? Thanks.

The situation is as follows. A solid disk of radius R and mass [itex]M_{w}[/itex] is resting on its edge on a horizontal plane surface near a rigid vertical pole. Free to slide frictionlessly on the pole is a collar of mass [itex]M_{c}[/itex]. Connecting the collar to a frictionless pin at the center of the disk is a thin beam of length [itex]L[/itex] and mass [itex]M_{b}[/itex]. The angle between the beam and the plane is [itex]\theta_{0}[/itex]. Assuming the system is released, find an expression for the velocity of the collar when the angle is θ.

To solve this I used conservation of mechanical energy.

The initial energy is all potential, and the energy at θ (once the system is released) is a mixture of kinetic and potential.

[itex]E_{{{\it before}}} = E_{{{\it pot.} {\it\space collar \space initial}}}+E_{{{\it pot.} {

\it beam \space initial}}}[/itex]

[itex]E_{{{\it pot.}{\it \space collar \space initial}}}=M_{{c}}\cdot g \cdot L\cdot \sin \left(

\theta_{{0}} \right)[/itex]

[itex]E_{{{\it pot.} {\it beam \space initial}}}=\frac{M_{{b}}\cdot g \cdot L\cdot \sin

\left( \theta_{{0}} \right)}{2}

[/itex]

Now we find the energies of the components when the system is released.

[itex]E_{{{\it kin.} {\it \space wheel}}}=\frac{{\it I}_{{{\it \space wheel}}}\cdot{\space

{\omega}^{2}_{{{\it wheel}}}}}{2}+\frac{M_{{w}}\cdot{{V}^{2}_{{c. m. {\it \space wheel}

}}}}{2}

[/itex]

[itex]\omega_{{{\it wheel}}}={\frac {V_{{c. m. {\it \space wheel}}}}{R}}[/itex]

[itex]{\it I}_{{{\it wheel}}}=\frac{M_{{w}}\cdot{R}^{2}}{2}[/itex]

[itex]V_{{c. m. {\it \space wheel}}}={\frac {V_{{c}}}{\tan \left( \theta

\right) }}

[/itex]

[itex]E_{{{\it kin.} {\it \space collar}}}=\frac{M_{{c}}{V_{{c}}}^{2}}{2}[/itex]

[itex]E_{{{\it pot.} {\it \space collar}}}=M_{{c}}\cdot g \cdot L\sin \left( \theta \right)[/itex]

[itex]E_{{{\it kin.}{\it \space beam}}}=\frac{{\it I}_{{c. m. {\it

\space beam}}}\cdot {{\omega}^{2}_{{{\it beam}}}}}{2} +\frac{M_{{b}}\cdot {{V}^{2}_{{c\cdot m\cdot {

\it beam}}}}

}{2}[/itex]

[itex]E_{{{\it pot.} {\it \space beam}}}=\frac{M_{{b}}\cdot g \cdot L\cdot \sin \left( \theta

\right)}{2}

[/itex]

[itex]{\it I}_{{c. m. {\it \space beam}}}=\frac{M_{{b}}{L}^{2}}{12}[/itex]

[itex]\omega_{{{\it beam}}}={\frac {2\cdot V_{{c}}\cdot \cos \left( \theta \right) }{L

}}

[/itex]

[itex]V_{{c. m.{\it \space beam}}}=\frac{V_{{c}}}{2}\cdot \sqrt {1+ \frac{1}{\left( {\tan}^{2}

\left( \theta \right) \right) }}

[/itex]

So putting this all together, we get an expression for the total energy of the system after release.

[itex]E_{{{\it after}}}=E_{{{\it kin.}{\it \space wheel}}}+E_{{{\it kin.}

{\it \space collar}}}+E_{{{\it pot.} {\it \space collar}}}+E_{{{\it kin.} {

\it \space beam}}}+E_{{{\it pot.}{\it \space beam}}}

[/itex]

Equating this with the initial energy, and solving for [itex]V_{c}[/itex] yields:

[itex]V_{c} = \sqrt { \frac{M_{{c}}\cdot g\cdot L\cdot \left( \sin \left( \beta \right) - \sin \left( \theta \right) \right)+\frac{M_{{b}}\cdot g\cdot L\cdot \left(\sin

\left( \beta \right) - \sin

\left( \theta \right) \right)}{2}}{{\frac {3\cdot M_{{w}}}{4\cdot

{\tan} ^{2} \left( \theta \right)}}+\frac{M_{{c}}}{2}+\frac{M

_{{b}}\cdot {\cos} ^{2} \left( \theta \right) }{6}+\frac{M_{{b}}

\cdot \left( 1+ \frac{1}{{\tan}^{2} \left( \theta \right) } \right)

}{8}}}

[/itex]

For definiteness, I use:

[itex]M_{w} = 21[/itex]

[itex]M_{c} = 33[/itex]

[itex]M_{b} = 11[/itex]

[itex]\theta_{0}=63°[/itex]

[itex]\theta=32°[/itex]

[itex]L = 3.11[/itex]

And get [itex]V_{c} = 2.593[/itex]

I found this problem while preparing for the qualifying exams and thought I would tackle it. I have no idea if it is correct, but I would like someone to tell me if there is an easier way to solve this than I used. Perhaps some Lagrangian or Hamiltonian approach? Thanks.

The situation is as follows. A solid disk of radius R and mass [itex]M_{w}[/itex] is resting on its edge on a horizontal plane surface near a rigid vertical pole. Free to slide frictionlessly on the pole is a collar of mass [itex]M_{c}[/itex]. Connecting the collar to a frictionless pin at the center of the disk is a thin beam of length [itex]L[/itex] and mass [itex]M_{b}[/itex]. The angle between the beam and the plane is [itex]\theta_{0}[/itex]. Assuming the system is released, find an expression for the velocity of the collar when the angle is θ.

To solve this I used conservation of mechanical energy.

The initial energy is all potential, and the energy at θ (once the system is released) is a mixture of kinetic and potential.

[itex]E_{{{\it before}}} = E_{{{\it pot.} {\it\space collar \space initial}}}+E_{{{\it pot.} {

\it beam \space initial}}}[/itex]

[itex]E_{{{\it pot.}{\it \space collar \space initial}}}=M_{{c}}\cdot g \cdot L\cdot \sin \left(

\theta_{{0}} \right)[/itex]

[itex]E_{{{\it pot.} {\it beam \space initial}}}=\frac{M_{{b}}\cdot g \cdot L\cdot \sin

\left( \theta_{{0}} \right)}{2}

[/itex]

Now we find the energies of the components when the system is released.

[itex]E_{{{\it kin.} {\it \space wheel}}}=\frac{{\it I}_{{{\it \space wheel}}}\cdot{\space

{\omega}^{2}_{{{\it wheel}}}}}{2}+\frac{M_{{w}}\cdot{{V}^{2}_{{c. m. {\it \space wheel}

}}}}{2}

[/itex]

[itex]\omega_{{{\it wheel}}}={\frac {V_{{c. m. {\it \space wheel}}}}{R}}[/itex]

[itex]{\it I}_{{{\it wheel}}}=\frac{M_{{w}}\cdot{R}^{2}}{2}[/itex]

[itex]V_{{c. m. {\it \space wheel}}}={\frac {V_{{c}}}{\tan \left( \theta

\right) }}

[/itex]

[itex]E_{{{\it kin.} {\it \space collar}}}=\frac{M_{{c}}{V_{{c}}}^{2}}{2}[/itex]

[itex]E_{{{\it pot.} {\it \space collar}}}=M_{{c}}\cdot g \cdot L\sin \left( \theta \right)[/itex]

[itex]E_{{{\it kin.}{\it \space beam}}}=\frac{{\it I}_{{c. m. {\it

\space beam}}}\cdot {{\omega}^{2}_{{{\it beam}}}}}{2} +\frac{M_{{b}}\cdot {{V}^{2}_{{c\cdot m\cdot {

\it beam}}}}

}{2}[/itex]

[itex]E_{{{\it pot.} {\it \space beam}}}=\frac{M_{{b}}\cdot g \cdot L\cdot \sin \left( \theta

\right)}{2}

[/itex]

[itex]{\it I}_{{c. m. {\it \space beam}}}=\frac{M_{{b}}{L}^{2}}{12}[/itex]

[itex]\omega_{{{\it beam}}}={\frac {2\cdot V_{{c}}\cdot \cos \left( \theta \right) }{L

}}

[/itex]

[itex]V_{{c. m.{\it \space beam}}}=\frac{V_{{c}}}{2}\cdot \sqrt {1+ \frac{1}{\left( {\tan}^{2}

\left( \theta \right) \right) }}

[/itex]

So putting this all together, we get an expression for the total energy of the system after release.

[itex]E_{{{\it after}}}=E_{{{\it kin.}{\it \space wheel}}}+E_{{{\it kin.}

{\it \space collar}}}+E_{{{\it pot.} {\it \space collar}}}+E_{{{\it kin.} {

\it \space beam}}}+E_{{{\it pot.}{\it \space beam}}}

[/itex]

Equating this with the initial energy, and solving for [itex]V_{c}[/itex] yields:

[itex]V_{c} = \sqrt { \frac{M_{{c}}\cdot g\cdot L\cdot \left( \sin \left( \beta \right) - \sin \left( \theta \right) \right)+\frac{M_{{b}}\cdot g\cdot L\cdot \left(\sin

\left( \beta \right) - \sin

\left( \theta \right) \right)}{2}}{{\frac {3\cdot M_{{w}}}{4\cdot

{\tan} ^{2} \left( \theta \right)}}+\frac{M_{{c}}}{2}+\frac{M

_{{b}}\cdot {\cos} ^{2} \left( \theta \right) }{6}+\frac{M_{{b}}

\cdot \left( 1+ \frac{1}{{\tan}^{2} \left( \theta \right) } \right)

}{8}}}

[/itex]

For definiteness, I use:

[itex]M_{w} = 21[/itex]

[itex]M_{c} = 33[/itex]

[itex]M_{b} = 11[/itex]

[itex]\theta_{0}=63°[/itex]

[itex]\theta=32°[/itex]

[itex]L = 3.11[/itex]

And get [itex]V_{c} = 2.593[/itex]

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