Rotating meterstick problem w/ weights.

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The problem involves a meter stick with identical particles at the 50-cm and 80-cm marks, pivoting at the 0-cm mark, and asks for the angular speed as it swings down. Initial attempts to solve using simultaneous equations and the parallel axis theorem were unsuccessful. A suggestion was made to apply conservation of energy, considering the potential and kinetic energy states of the meter stick. This approach focuses on the energy transformation as the stick moves from a horizontal to a vertical position. Utilizing conservation of energy is deemed a more effective method for solving the problem.
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Homework Statement


Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate about a pivot at the 0-cm mark on the meter stick. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through it's lowest position?

Homework Equations


Equations of rotational motion
torque = Fdsin(theta)
torque = I (alpha)
v = rw
(alpha) = A_c / r

The Attempt at a Solution


I used the different lengths in simultaneous equations to try and find the different omegas and add them together...but my answer had nothing to do with the question...

I then tried using parallel axis theorem and tried to plug in T = I(alpha) and such to figure out the real torque involved to find alpha and find w finally., and that produced no results...
 
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jrrodri7 said:

Homework Statement


Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate about a pivot at the 0-cm mark on the meter stick. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through it's lowest position?

Homework Equations


Equations of rotational motion
torque = Fdsin(theta)
torque = I (alpha)
v = rw
(alpha) = A_c / r

The Attempt at a Solution


I used the different lengths in simultaneous equations to try and find the different omegas and add them together...but my answer had nothing to do with the question...

I then tried using parallel axis theorem and tried to plug in T = I(alpha) and such to figure out the real torque involved to find alpha and find w finally., and that produced no results...
Perhaps this problem would be better approached using conservation of energy.
 
conservation of rotating object energy?hmmmm okay.
 
jrrodri7 said:
conservation of rotating object energy?hmmmm okay.
Yes. You don't think that one can apply conservation of energy to rotating rigid bodies? Consider the potential and kinetic energy of the meter stick both when it is horizontal and vertical.
 
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