- #1
kof9595995
- 679
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Yesterday my friend asked me: what will happen if a force F with constant magnitude is applied to one end of a free stick(no other force), and this force F is always normal to the stick.
He told me it's not a textbook problem and he just asked out of curiosity,so I decide not to put this thread in homework section, but still I'll show my attempt to the solution.
Suppose the stick is uniform with length l and mass M
Let's denote the position of center of mass as
[tex]\[\overrightarrow {{r_{cm}}} = (x,y)\][/tex]
The angle rotates over is [tex]\[\theta \][/tex]
then the point that the force applied at is
[tex]\[\overrightarrow {{r_F}} = (x - \frac{l}{2}\cos \theta ,y + \frac{l}{2}\sin \theta )\][/tex]
I choose the axis of rotation to be the axis perpendicular to the paper and passing thru the origin,and we know force is
[tex]\[\overrightarrow F = (F\sin \theta ,F\cos \theta )\][/tex]
then I can express torque as
[tex]\[\overrightarrow \tau = \overrightarrow {{r_F}} \times \overrightarrow F = F(x\cos \theta - y\sin \theta - \frac{l}{2})\hat k\][/tex]
About the moment of inertia I apply the parallel axis theorem
[tex]\[I = {I_{cm}} + M{D^2} = \frac{1}{{12}}M{l^2} + M({x^2} + {y^2})\][/tex]
Now we can just apply Newton's law:
[tex]\[\left\{ \begin{array}{l}
\ddot x = \frac{F}{M}\sin \theta \\
\ddot y = \frac{F}{M}\cos \theta \\
\ddot \theta = \frac{{|\overrightarrow \tau |}}{I} = \frac{F}{M}\frac{{x\cos \theta - y\sin \theta - \frac{l}{2}}}{{\frac{{{l^2}}}{{12}} + {x^2} + {y^2}}} \\
\end{array} \right.\][/tex]
Then I'm totally stuck, I'm not that optimistic to expect an exact solution but the embarrassing thing is I can't even reduce the system to a single variable diff equation.
Can anybody help me reduce this? Any help is appreciated. Of course if somebody tells me there's an exact solution, I‘ll be happy enough to have a party.
(I checked these equations one time and hopefully no mistake. )
He told me it's not a textbook problem and he just asked out of curiosity,so I decide not to put this thread in homework section, but still I'll show my attempt to the solution.
Suppose the stick is uniform with length l and mass M
Let's denote the position of center of mass as
[tex]\[\overrightarrow {{r_{cm}}} = (x,y)\][/tex]
The angle rotates over is [tex]\[\theta \][/tex]
then the point that the force applied at is
[tex]\[\overrightarrow {{r_F}} = (x - \frac{l}{2}\cos \theta ,y + \frac{l}{2}\sin \theta )\][/tex]
I choose the axis of rotation to be the axis perpendicular to the paper and passing thru the origin,and we know force is
[tex]\[\overrightarrow F = (F\sin \theta ,F\cos \theta )\][/tex]
then I can express torque as
[tex]\[\overrightarrow \tau = \overrightarrow {{r_F}} \times \overrightarrow F = F(x\cos \theta - y\sin \theta - \frac{l}{2})\hat k\][/tex]
About the moment of inertia I apply the parallel axis theorem
[tex]\[I = {I_{cm}} + M{D^2} = \frac{1}{{12}}M{l^2} + M({x^2} + {y^2})\][/tex]
Now we can just apply Newton's law:
[tex]\[\left\{ \begin{array}{l}
\ddot x = \frac{F}{M}\sin \theta \\
\ddot y = \frac{F}{M}\cos \theta \\
\ddot \theta = \frac{{|\overrightarrow \tau |}}{I} = \frac{F}{M}\frac{{x\cos \theta - y\sin \theta - \frac{l}{2}}}{{\frac{{{l^2}}}{{12}} + {x^2} + {y^2}}} \\
\end{array} \right.\][/tex]
Then I'm totally stuck, I'm not that optimistic to expect an exact solution but the embarrassing thing is I can't even reduce the system to a single variable diff equation.
Can anybody help me reduce this? Any help is appreciated. Of course if somebody tells me there's an exact solution, I‘ll be happy enough to have a party.
(I checked these equations one time and hopefully no mistake. )