Calculating Speed and Kinetic Energy in a Rotating Rod

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SUMMARY

The discussion focuses on calculating the speed and kinetic energy of a rotating rod with a mass of 0.380 kg and a length of 2.50 m, pivoted at one end. For part (a), the correct method involves using the moment of inertia formula (1/3MR²) and energy conservation principles to find the speed of the center of gravity at its lowest position. For part (b), the tangential speed of the free end of the rod is derived from the angular velocity calculated in part (a), not simply doubled. The initial calculations presented were incorrect due to misapplication of energy principles and failure to account for rotational dynamics.

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Homework Statement


[/B]
The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.380 kg and its length is 2.50 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown.

(a) Determine the speed of its center of gravity at its lowest position.

(b) When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod.

Homework Equations



Moment of Inertia for a solid rod: 1/3MR^2

Tangential Acceleration = r * (dω/dt)

The Attempt at a Solution



I really doubt this is right, but this is what I have been trying.

For part A, this is what I did.
To get the center of mass, I tried 2.5/2 for 1.25.
For the speed at that point, I tried using mgh for potential/kinetic energy and got (0.38g)(9.81m/s^2)(1.25). This gave me 4.66. Then I square rooted 4.66 over 0.38 (the mass), and got 4.95.

My answer was incorrect

For part B, I assumed that it was the answer for part A multiplied by two, since A was asking for the center, rather than the full rod. I'll need to figure out part A before part B.
 
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tamakitty said:
I really doubt this is right, but this is what I have been trying.

For part A, this is what I did.
To get the center of mass, I tried 2.5/2 for 1.25.
For the speed at that point, I tried using mgh for potential/kinetic energy and got (0.38g)(9.81m/s^2)(1.25). This gave me 4.66. Then I square rooted 4.66 over 0.38 (the mass), and got 4.95.

My answer was incorrect
You should always show the units for the values you calculate. For example, what units are associated with the 4.66 value?

Since the mass was not falling linearly but rotationally, what form of kinetic energy is involved? What's an expression for it?
 

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