Rod w/ Pivot and Attached Mass Pendulum

[srekai]

Homework Statement

A solid rod of length L and mass M has a pivot through its center and is originally horizontal. Another mass 2M is then attached firmly to one end of the rod, and released. What is the maximum speed of the mass 2M attained thereafter? (Cornell 2009)

Homework Equations

Not 100% sure, see solution below
Approach 1:
PE = mgh
KE = $\frac{1}{2}mv^2$

Approach 2:
Torque: $\tau = rFsin \theta$
Moment of Inertia: $I = mr^2$
Torque: $\tau = I \alpha$

The Attempt at a Solution

I approached this two ways, first as a potential energy problem via a pendulum.
The max velocity of this object will be achieved when the pivot is 90 degrees.
So I set the half of the arm with the mass to be a pendulum with mass $\frac{5M}{2}$
The starting height of the pendulum will be half of the rod's length $\frac{L}{2}$

Setting KE = PE
$$\frac{1}{2} \cdot \frac{5M}{2} \cdot v^2 = \frac{5M}{2} \cdot g \cdot \frac{L}{2}$$
$$v = \sqrt{gL}$$

This doesn't seem quite right to me of course, so I tried to approach this as a problem with torque
Setting torque equal to $\tau = \frac{L}{2} \frac{5M}{2} \cdot g \cdot sin(\theta)$
Then moment of Inertia is $I = \frac{5M}{2} * \frac{L}{2}^2$

Not sure how to clear the rest of this approach.

 

[kuruman]

Energy conservation is the way to go because the acceleration is not constant so you cannot use the standard kinematic equations. For the kinetic energy you should use the rotational form which means you need to find the moment of inertia about the pivot. You also need to find where the center of mass of this thing is and find how far it drops. That should be the $y$ in $mgy$.

 

[srekai]

So, am I correct in stating that the moment of inertia is $I = mr^2 = \frac{5M}{2} \cdot \frac{L}{2}^2$?

Then, I would solve for KE = PE, where KE = $\frac{1}{2} I \omega^2 = mgy =$ PE

Thereby, I solve for $\omega$ as $\omega = \sqrt{\frac{2mgy}{I}}$, and then v = $r \omega = \sqrt{2gy}$

My main question now is, how would I calculate y? I tried using the center of mass equation $m_1 r_1 = m_2 r_2$, and this means that $r_1 = 5r_2$. Not sure how to go beyond this.

Supposing I use the CoM equation as this $x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1+m_2}$, with $x_1$ as 0, and $x_2$ as L, I can set $x_{cm} = \frac{5}{6} L$, is this my value?

 

[person123]

I actually don't think you need to find the center of mass of the entire system. Does the center of mass of the rod (without the attached mass) change?

 

[haruspex]

I actually don't think you need to find the center of mass of the entire system

Yes, it is almost always easier to deal with moments of inertia, momenta, energy... separately for the simple components rather than as a compound body.