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
Andy Y.
Feb 9, 2017

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

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