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,

Last edited: Mar 22, 2005
5. Mar 22, 2005

flower76

The equation part I'm ok with, but the second question is what is giving me trouble, could anyone confirm that I'm right in saying that the maximum angular velocity occurs when the rod is vertical? Or am I totally wrong?

Thanks