# What is the final angular speed of the barrel? -->

## A barrel is given a downhill rolling start of $1.5 \text{ rad/s}$ at the top of a hill. Assume a constant angular acceleration of $2.9 {\text{ rad/s}}^{2}$ If the barrel takes $11.5 \text{ s}$ to get to the bottom of the hill, what is the final angular speed of the barrel? What angular displacement does the barrel experience during the $11.5 \text{ s}$ ride?

Mar 15, 2018

Part 1

Begin with the equation:

$\omega = {\omega}_{0} + \alpha t$

where $\omega$ is the final angular speed, ${\omega}_{0}$ is initial angular speed, $\alpha$ is the angular acceleration, and $t$ is the time of the acceleration.

Given: ${\omega}_{0} = 1.5 \text{ rad/s}$, $\alpha = 2.9 {\text{ rad/s}}^{2}$, and $t = 11.5 \text{ s}$

omega = 1.5" rad/s" + (2.9" rad/s"^2)(11.5" s")

$\omega = 38.45 \text{ rad/s}$

Part 2

Begin with the equation:

$\theta = {\theta}_{0} + {\omega}_{0} t + \frac{1}{2} \alpha {t}^{2}$

We are not given ${\theta}_{0}$, therefore, we assume that it is 0:

$\theta = {\left(1.5 \text{ rad/s")(11.5" s") + 1/2(2.9" rad/s"^2)(11.5" s}\right)}^{2}$

$\theta = 209 \text{ rad}$