# Angular velocity of a rod - Help!

1. Mar 21, 2005

### flower76

I have figured out the first part regarding the equation, but I don't really know what to write for the second part. Intuitively I think the rod would swing down, because m1 is heavier, and it would have the greatest velocity when vertical. But I have no idea if I'm right, and if I am I can't give reasons why.

The question:

A rigid rod of mass M and length I, with masses m1 and m2 attached at the end of the rod, can rotate in a vertical plane about a frictionless pivot through its centre. At time t=0, the rod is held at an angle theta, as shown. Show that if the system is let go, the angular acceleration of the rod at t=0 is:

$$\alpha=\frac{2(m1-m2)gcos\theta}{L(M/3+m1+m2)}$$

If m1>m2, for what value of $$\theta$$ is the angular velocity $$\omega$$ a maximum? Give reasons for your answer.

The diagram shown has a rod at about a 45 degree angle north east with m1 the mass on the bottom.

Help is much appreciated.

2. Mar 22, 2005

### Timbuqtu

I assume you know the formula:

$$\tau = I \alpha$$

where $$\tau$$ is the torque, $$I$$ is the moment of inertia around the axis and $$\alpha = d\omega/dt = d^2\theta/dt^2$$ is the angular acceleration.

The moment of inertia of the rod is $$1/12 M L^2$$, so:

$$I = 1/12 M L^2 + m_1 (L/2)^2 + m_2 (L/2)^2$$

and the torque due to the gravitational force is the sum of the torques on both mass m1 and m2, but one of them with a minus sine because they point in opposite directions.

Concerning the other question: In this case total energy is conserved and kinetic energy is proportional to $$\omega^2$$, so what can you say about the potential energy when $$\omega$$ is maximal?

3. Mar 22, 2005

### flower76

So when angular velocity is at a maximum, potential energy will be at a minimum, meaning that the rod will be vertical??

Is this correct?

4. Mar 22, 2005

Ill give you a hand anf give you a head start on the first question. To prrove the angular acceleration.
First, draw a free body diagram of the situation. You will see that the sum of the torque about the centre point is:
$$\Sigma \tau = \frac{L}{2} cos \theta m_{1} g - \frac{L}{2} cos \theta m_{2} g$$

we can factor out a term:

$$\Sigma \tau = \frac{L}{2} cos \theta g (m_1 - m_2)$$

now, we alson know that:
$$\Sigma \tau = I\alpha$$

the moment of inertia for the system is the sum of all of the moments of inertia.
$$\Sigma I = I_{rod} + I_{m_1} + I_{m_2}$$

$$\Sigma I = \frac{1}{12}ML^2 + m_1 (\frac{L}{2})^2 + m_2 (\frac{L}{2})^2$$

$$\Sigma I = \frac{1}{12}ML^2 + m_1 \frac{L^2}{4} + m_2 \frac{L^2}{4}$$

$$\Sigma I = \frac{L^2}{4} (\frac{1}{3} M + m_1 + m_2)$$

I think you should be able to do the rest by yourself. If you still don't understand, ask.

Regards,