Finding Angular Acceleration of Uniform Rod

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Homework Help Overview

The problem involves a uniform rod pivoted at one end, released from a vertical position, and the task is to find its angular acceleration and the components of acceleration of its center of mass when the rod is horizontal. The subject area pertains to rotational dynamics and angular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate angular acceleration using angular velocity and time, but expresses uncertainty about how to determine the time. Some participants question the validity of using the equation for angular acceleration due to the lack of information about uniformity and time.

Discussion Status

Participants are exploring different approaches to the problem, with some suggesting the use of torque and moment of inertia instead of the original method proposed by the poster. There is an ongoing inquiry into how to calculate the components of acceleration for the center of mass, indicating a productive direction in the discussion.

Contextual Notes

The original poster mentions specific numerical values for the correct answers to parts (a) and (b), but the discussion focuses on the methods and reasoning rather than confirming these results.

roam
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Homework Statement



A uniform rod of length 0.750 m and mass 1.20 kg is pivoted at one end by a smooth pin as shown below. The rod is released from the vertical position and given a slight nudge to release it from the vertical position of unstable equilibrium.

[PLAIN]http://img175.imageshack.us/img175/9820/angacc40pc.png

When the rod is horizontal:

(a) Calculate its angular acceleration.

(b) Calculate the x and the y components of the acceleration of the centre of mass.

The Attempt at a Solution



First I focus on part (a). I have already obtained the angular velocity:

The potential energy of the system relative to the reference configuration is MgL/2 because the center of mass of the rod is at a height L/2 away from its position in the reference configuration. Conservation of mechanical energy for the system is:

Ki+Ui = Kf + Uf

0+½MgL=½Iω²+0

[tex]\omega = \sqrt{\frac{MgL}{\frac{1}{3}ML^2}} = \sqrt{\frac{3g}{L}}[/tex]

using the given values I get ω=6.26 rad/s, which is the correct velocity.

Now I want to use the equation [tex]\alpha = \frac{\Delta \omega}{t}[/tex] to find the angular acceleration. Here's where I'm stuck. Could anyone please show me how to find the time for this equation?

Correct answer to part (a) is 19.62 rad/s² and for part (b) it is -14.7 m/s²(x-component) and -7.36 m/s² (y-component)
 
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No, you can not use

[tex] \alpha = \frac{\Delta \omega}{t}[/tex],
as it is valid for uniform acceleration only, and you do not know if the acceleration is uniform; neither you know the time.

There is a formula betwee torque, angular acceleration, and moment of inertia. Use that.

ehild
 
ehild said:
No, you can not use

[tex] \alpha = \frac{\Delta \omega}{t}[/tex],
as it is valid for uniform acceleration only, and you do not know if the acceleration is uniform; neither you know the time.

There is a formula betwee torque, angular acceleration, and moment of inertia. Use that.

ehild

Thank you very much, I got it. Now could you explain how to deal with part (b)? How do we calculate the x (or y) component of the acceleration of the centre of mass? I'm not quite sure what that even means since I don't have any worked examples in my textbook.
 
The centre of mass moves along a circle, and the angular velocity is not constant. What components of acceleration has a body performing circular motion?

ehild
 

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