How Is Gravitational Potential Energy Calculated in a Rotating Stick Scenario?

[Kenchin]

A stick with a mass of 0.170Kg and a length of 1.00m is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released.

1) As it swings through the vertical, calculatethe change in gravitational potential energy that has occurred. Gravity = 9.81m/s^2

Alright So for this one I have no idea really where to begin except to find the Moment of inertia. 1/3 ML^2. Substituting I get 1/3 (.170kg) (1)^2

The problem with that is is that I don't know how to find the potential energy because there is no height given to use U=mgh. I tried to use Mgy(y is in cm) and that didnt work well because it gave me (.170kg)(9.81m/s^2)(100cm). Where would I go from here considering there is no w (angular velocity) or anything given.

 

[Hootenanny]

You are correct, you don't know the actual height, but you do know the change in height. Consider the centre of mass of the rule, a sketch may be helpful.

You don't actually need any moment of inertia calculations for this question.

-Hoot

 

[Kenchin]

The problem I guess that I'm having is visualizing the problem, the way I visualize it is a pendulum starting from the 0 or 2(pi) section and swinging to the (pi). Is this the proper visualization? If so then the maximum height would be 2m's. Then in this case it would be KE_1+PE_1=KE_2+PE_2. This way, KE_1=0, PE_1=.170Kg(9.81 m/s^2)(2m). But if this was the case then all the potential energy would convert into kinetic energy at the bottom of the swing (through the verticle). But this then yields nothing helpful, what might be wrong with my visualization?

 

[Hootenanny]

The question states that the rule begins horizontal, the rule is then release and allowed to swing freely. The question asks what the change in potential energy is when the rule is pointing vertically down. Imagine the rule is a straight horiztonal line at y = 0, the rule now pivots about the origin until it is a straight line at x = 0. Think about that displacement of the centre of mass.

-Hoot