What torque would have to be applied to a rod with a length of #3 m# and a mass of #6 kg# to change its horizontal spin by a frequency #5 Hz# over #9 s#?

1 Answer
Feb 9, 2017

Answer:

#tau_"net"=5/6("kg m")/s^2~~0.833("kg m")/s^2#

Explanation:

Assuming the rod in question rotates about the center (not the end), we know that

Step 1. Gather the information that you know and need

Mass of rod: #"mass"=m=6" kg"#
Length of rod: #"length"=L=3" m"#
Change in angular speed: #domega=5" Hz"#
Change in time: #dt=9" s"#
Torque required: ??

Step 2. Determine the formula using the above givens

Torque: #tau_"net"=Ialpha#

  • #I# is the moment of inertia of our rod
  • Assuming center spin, #I_"rod"=1"/"12*m*L#
  • #alpha# is the instantaneous angular acceleration
  • #alpha=domega"/"dt# with units #"rad/"s^2# or #s^-2#

Torque: #tau_"net"=1/12mL (domega)/dt#

Step 3. Plug your knows into the formula for torque

#tau_"net"=1/12(6" kg")(3" m")(5" Hz")/(9" s")=5/6("kg m")/s^2~~0.833("kg m")/s^2#