- #1
Radarithm
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Edit: Can someone change the name of the thread somehow? I accidentally posted it without changing the name.
(Moderator note -- title updated.)
The question is quite long so here is a picture: http://gyazo.com/dc917d1885b6ffebb0a39e2409af4d61
$$E=T+V$$
$$T=\frac{m\dot{x}^2}{2}$$
$$U_g=mgh$$
$$\Gamma=I\alpha$$
If we set the zero of potential energy at origin (where the mass is at when the system is in equilibrium), then the change in gravitational potential is:
$$U(\phi)=mg(l-l\cos{\phi})=mgl(1-\cos{\phi})$$
Now that we're done with part (a), I move on to part (b), which is where I think I messed up.
The total mechanical energy of the system can be defined to be:
$$E(\phi, \dot{\phi})=T(\phi,\dot{\phi})+U(\phi)$$
Where the Kinetic energy (T) is:
$$T(\phi,\dot{\phi})=\frac{m\dot{\phi}l\sin{\phi}}{2}$$
and the potential, U, is:
$$U(\phi)=mgl(1-\cos{\phi})$$
Here is where I think I messed up mathematically; taking the time derivative of the mechanical energy, I get this:
$$\dot{E}=ml\left(\frac{\dot{\phi}\ddot{\phi}\cos{\phi}}{2}+g(\dot{\phi}\, \cos{\phi}+1)\right)$$
Did I go wrong somewhere?
(Moderator note -- title updated.)
Homework Statement
The question is quite long so here is a picture: http://gyazo.com/dc917d1885b6ffebb0a39e2409af4d61
Homework Equations
$$E=T+V$$
$$T=\frac{m\dot{x}^2}{2}$$
$$U_g=mgh$$
$$\Gamma=I\alpha$$
The Attempt at a Solution
If we set the zero of potential energy at origin (where the mass is at when the system is in equilibrium), then the change in gravitational potential is:
$$U(\phi)=mg(l-l\cos{\phi})=mgl(1-\cos{\phi})$$
Now that we're done with part (a), I move on to part (b), which is where I think I messed up.
The total mechanical energy of the system can be defined to be:
$$E(\phi, \dot{\phi})=T(\phi,\dot{\phi})+U(\phi)$$
Where the Kinetic energy (T) is:
$$T(\phi,\dot{\phi})=\frac{m\dot{\phi}l\sin{\phi}}{2}$$
and the potential, U, is:
$$U(\phi)=mgl(1-\cos{\phi})$$
Here is where I think I messed up mathematically; taking the time derivative of the mechanical energy, I get this:
$$\dot{E}=ml\left(\frac{\dot{\phi}\ddot{\phi}\cos{\phi}}{2}+g(\dot{\phi}\, \cos{\phi}+1)\right)$$
Did I go wrong somewhere?
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