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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?

1 Answer
Mar 15, 2018

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"#