- #1

srekai

- 8

- 0

## 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.