Rotating meterstick problem w/ weights.

  • Thread starter jrrodri7
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In summary: The potential energy of the meter stick is the distance from the pivot to the bottom of the swing. The kinetic energy of the meter stick is the speed of the meter stick at the bottom of the swing. The total potential and kinetic energy is the same when the meter stick is horizontal and vertical.
  • #1
jrrodri7
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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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  • #2
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.
 
  • #3
conservation of rotating object energy?hmmmm okay.
 
  • #4
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.
 
Last edited:

1. What is the purpose of the rotating meterstick problem with weights?

The purpose of the rotating meterstick problem with weights is to demonstrate the principles of rotational equilibrium and torque. It involves a meterstick attached to a pivot point with weights hanging from different positions on the meterstick, and the goal is to find the position of the pivot point where the meterstick will remain balanced and not rotate.

2. How is torque calculated in the rotating meterstick problem?

Torque is calculated by multiplying the weight of an object by its distance from the pivot point. In the rotating meterstick problem, the torque of each weight is calculated and then compared to determine the position of the pivot point where the torques are balanced.

3. What factors affect the balance of the rotating meterstick?

The balance of the rotating meterstick is affected by the weight of the objects, their distances from the pivot point, and the angle at which they hang from the meterstick. These factors determine the torque of each weight and therefore impact the overall balance of the meterstick.

4. How does the center of mass relate to the rotating meterstick problem?

The center of mass, or the point at which the weight of an object is evenly distributed, is an important concept in the rotating meterstick problem. The pivot point must be positioned directly below the center of mass in order for the meterstick to remain balanced.

5. What are some real-life applications of the rotating meterstick problem?

The rotating meterstick problem has real-life applications in fields such as engineering, physics, and architecture. It can help determine the optimal placement of objects in a structure to ensure balance and stability, as well as calculate the torque and forces acting on different parts of a system.

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