Question

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

A barrel is given a downhill rolling start of `1.5" rad/s"` at the top of a hill. Assume a constant angular acceleration of `2.9" rad/s"^2`

1. If the barrel takes `11.5" s"` to get to the bottom of the hill, what is the final angular speed of the barrel?

2. What angular displacement does the barrel experience during the `11.5" s"` ride?

Answer 1

Part 1

Begin with the equation:

`omega = omega_0+alphat`

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" rad/s"`, `alpha = 2.9" rad/s"^2`, and `t = 11.5" s"`

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

`omega = 38.45" rad/s"`

Part 2

Begin with the equation:

`theta = theta_0+ omega_0t+1/2alphat^2`

We are not given `theta_0`, therefore, we assume that it is 0:

`theta = (1.5" rad/s")(11.5" s") + 1/2(2.9" rad/s"^2)(11.5" s")^2`

`theta = 209" rad"`