## Rotation about a fixed axis

1. The problem statement, all variables and given/known data

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?

2. Relevant 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: $$\sum \tau = I \alpha$$

3. The attempt at a solution

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

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

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

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 Quote by roam 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 $$\alpha = 31.6$$...

## Rotation about a fixed axis

Typing error

torque=I*alpha
torque=Fr

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 Quote by roam 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?

 Quote by Doc Al 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

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 Quote by inky distance=0.1 cm
The question was meant for the OP, of course.