# 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?

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: $\text{mass"=m=6" kg}$
Length of rod: $\text{length"=L=3" m}$
Change in angular speed: $\mathrm{do} m e g a = 5 \text{ Hz}$
Change in time: $\mathrm{dt} = 9 \text{ s}$
Torque required: ??

Step 2. Determine the formula using the above givens

Torque: ${\tau}_{\text{net}} = I \alpha$

• $I$ is the moment of inertia of our rod
• Assuming center spin, ${I}_{\text{rod"=1"/}} 12 \cdot m \cdot L$
• $\alpha$ is the instantaneous angular acceleration
• $\alpha = \mathrm{do} m e g a \text{/} \mathrm{dt}$ with units $\text{rad/} {s}^{2}$ or ${s}^{-} 2$

Torque: ${\tau}_{\text{net}} = \frac{1}{12} m L \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

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