Calculating Angular Acceleration of a Rotating Rod

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A uniform metre-rule pivoted at the 40-cm mark is analyzed for angular acceleration after being released from a horizontal position. The torque due to gravitational force is calculated using the formula τ = Mg(L/2), with L being the length of the rod. The initial calculation for angular acceleration yielded an incorrect result, prompting a discussion on the correct distance from the pivot to the center of mass, which is 0.1 m. Clarifications on torque calculations emphasize the importance of accurate distance measurements and unit consistency. Ultimately, the correct angular acceleration is determined to be 10.5 rad/s².
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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 \mu kg/m and its rotational inertia about this pivot is 0.093 \mu 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: \tau =Mg \left( \frac{L}{2} \right)

Angular accleration and torque: \sum \tau = I \alpha

The Attempt at a Solution



Since \mu = \frac{M}{L}, and L=0.4m I get

\tau = 0.4 \mu g \frac{0.4}{2} = 0.78 \mu

\alpha = \frac{\tau}{I} = \frac{0.78 \mu}{0.093 \mu} = 8.43

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: \tau =Mg \left( \frac{L}{2} \right)
Why L/2?

Since \mu = \frac{M}{L}, 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 \alpha = 31.6...
 
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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