Angular Acceleration of Rod w/ Pivoted Weight

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Homework Help Overview

The problem involves a uniform rod pivoted at its center with a weight attached to one end. The scenario is set up to analyze the angular acceleration of the system when released from a specific angle, with given parameters such as the mass of the rod, the mass of the weight, and the length of the rod.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy, questioning the definitions and values used in the calculations, including the moment of inertia and the change in height of the weight.

Discussion Status

Participants are actively engaging in clarifying the variables involved in the energy equations and the physical setup. Some guidance has been provided regarding the calculation of height changes and the potential energy of the system, but no consensus has been reached on the correct approach or values.

Contextual Notes

There is an emphasis on understanding the change in height of the weight as the rod moves from an angle of 37 degrees to vertical, with some participants noting that the potential energy of the rod does not change due to its pivot point being at the center of mass.

mrnastytime
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Homework Statement


A uniform rod is pivoted at its center and a small weight of mass M = 5.07 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod.


Homework Equations


Now suppose that the rod has length L = 5.4 m and mass mrod = 10.1 kg. Suppose also that there are no external forces applied (i.e. Fh = Fv = 0). The system is released from rest at the q = 37° angle. What is the angular acceleration just after it is released?
angular acceleration:1.7437 rad/s^2
What is the angular velocity when the rod is vertical?

The Attempt at a Solution


mgh=.5Iw^2
where h=sin 53*L/2=2.156
w=sqrt(2mgh/I)
this is not the correct answer. Can somebody help me. I don't know what I am doing wrong.
 
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mrnastytime said:

The Attempt at a Solution


mgh=.5Iw^2
OK. (What's I? What's m?)
where h=sin 53*L/2=2.156
What's h the change in height of?
 
sorry i don't quite understand what you're asking, I=moment of inertia m=mass w=omega
the potential energy is mgh and kinetic energy is .5Iw^2
 
I meant: I = moment of inertia of what? m = mass of what? h = change in height of what?
 
I=moment of inertia of rod-weight system=61.50
m=10.1 kg
h=change in height of the rod from 37 degrees to 90 degrees
 
mrnastytime said:
I=moment of inertia of rod-weight system=61.50
OK.
m=10.1 kg
That's just the mass of the rod.
h=change in height of the rod from 37 degrees to 90 degrees
What part of the rod? What's the change in height as it moves from one position to the other?

How does the gravitational PE of the entire system change? What's the change in PE of the rod? Of the attached weight?
 
The part of the rod that's attached to the weight. The change in height is the point from which theta=37 degrees to the height when the rod is vertical at 90 degrees. The potential energy of the rod does not change because the pivot point is the center of mass?
 
mrnastytime said:
The part of the rod that's attached to the weight. The change in height is the point from which theta=37 degrees to the height when the rod is vertical at 90 degrees. The potential energy of the rod does not change because the pivot point is the center of mass?
All good. Redo your calculation of the change in height of the attached weight. (It's not L/2*sin(53).)
 
im having trouble redoing the calculation for the change in height. is the sin 53 part correct? because if you want the rod to be vertical, the tip of the rod with the weight will be pointing down, thus 90-37=53 degrees?
 
  • #10
Do this: Find the initial position of the tip of the rod (measured from the pivot point). Find the final position. Compare.
 

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