Calculating Angular Acceleration of a Rotating Rod

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Homework Help Overview

The problem involves calculating the angular acceleration of a uniform metre-rule pivoted at the 40-cm mark. The rod is released from a horizontal position, and participants are discussing the torque due to gravitational force and its relation to angular acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the torque and angular acceleration using the given mass distribution and pivot point. Questions are raised regarding the correct distance from the pivot for calculating torque and the reasoning behind using L/2 for the center of mass.

Discussion Status

There is an ongoing exploration of the correct application of torque and the distances involved. Some participants are questioning the assumptions made about the center of mass and the distances used in calculations, while others are clarifying the implications of these factors on the results.

Contextual Notes

Participants note potential errors in unit conversions and the need for clarity on the distance from the pivot to the center of mass, which is crucial for calculating torque accurately.

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Homework Statement



A uniform metre-rule is pivoted to rotate about a horizontal axisthrough the 40-cm mark. The stick has a mass/unit length of [tex]\mu[/tex] kg/m and its rotational inertia about this pivot is [tex]0.093 \mu[/tex] kg/m2. It is released from rest in a horizontal position. What is the magnitude of the angular acceleration of the rod?

Homework Equations



An expression for the magnitude of the torque due to gravitational force about an axis through the pivot: [tex]\tau =Mg \left( \frac{L}{2} \right)[/tex]

Angular acceleration and torque: [tex]\sum \tau = I \alpha[/tex]

The Attempt at a Solution



Since [tex]\mu = \frac{M}{L}[/tex], and L=0.4m I get

[tex]\tau = 0.4 \mu g \frac{0.4}{2} = 0.78 \mu[/tex]

[tex]\alpha = \frac{\tau}{I} = \frac{0.78 \mu}{0.093 \mu} = 8.43[/tex]

But my answer is wrong. The correct answer must be 10.5 rad/s2. Could anyone please help me? :(
 
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roam said:
An expression for the magnitude of the torque due to gravitational force about an axis through the pivot: [tex]\tau =Mg \left( \frac{L}{2} \right)[/tex]
Why L/2?

Since [tex]\mu = \frac{M}{L}[/tex], and L=0.4m I get
L = 1 m. (It's a meterstick.)

What force provides the torque? How far does that force act from the pivot?
 
Why L/2?

Because the gravitational force on the stick acts at its center of mass.

Since the it is pivoted at the 40 cm mark, should I use 60/2 (since the mass is uniformly distributed)?

But this didn't work because I ended up with [tex]\alpha = 31.6[/tex]...
 
Typing error

torque=I*alpha
torque=Fr
 
Last edited:
roam said:
Because the gravitational force on the stick acts at its center of mass.
True, but to calculate the torque due to the weight you need its distance from the pivot. What is that distance?
 
Doc Al said:
True, but to calculate the torque due to the weight you need its distance from the pivot. What is that distance?

distance=0.1 cm
 
inky said:
distance=0.1 cm
The question was meant for the OP, of course. :rolleyes:
(And your units are off.)
 
Doc Al said:
The question was meant for the OP, of course. :rolleyes:
(And your units are off.)

I am sorry for my wrong units. Actually I should write r=0.1 m. Thanks so much.
 

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