Question

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`?

Answer 1

`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`