1. Jun 17, 2010

roam

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

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

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

2. Jun 17, 2010

Staff: Mentor

Why L/2?

L = 1 m. (It's a meterstick.)

What force provides the torque? How far does that force act from the pivot?

3. Jun 18, 2010

roam

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 $$\alpha = 31.6$$...

4. Jun 18, 2010

inky

Typing error

torque=I*alpha
torque=Fr

Last edited: Jun 18, 2010
5. Jun 18, 2010

Staff: Mentor

True, but to calculate the torque due to the weight you need its distance from the pivot. What is that distance?

6. Jun 18, 2010

inky

distance=0.1 cm

7. Jun 18, 2010

Staff: Mentor

The question was meant for the OP, of course.