# What torque would have to be applied to a rod with a length of #4 m# and a mass of #3 kg# to change its horizontal spin by a frequency of #2 Hz# over #3 s#?

##### 1 Answer

I got

The **sum of the torques** for a rod is written as:

#\mathbf(sum vectau = Ibaralpha)#

I assume we are looking at a rod rotating horizontally on a flat surface (so that no vertical forces apply), **changing in angular velocity over time**.

Then, we are looking at the **average** angular acceleration

#\mathbf(baralpha = (Deltavecomega)/(Deltat) = (vecomega_f - vecomega_i)/(t_f - t_i) = (vecomega_f)/t),# where if we assume the

initial angular velocity#vecomega_i = 0# , then#vecomega_f = Deltavecomega# , and withinitial time#t_i = 0, t_f = t# .

Next, for this ideal rod of length **moment of inertia** *center of mass*, we need to take the integral for a uniform rod rotating about its center of mass (the midpoint

#color(green)(I_"cm") = m/Lint_(-L/2)^(L/2) r^2dr# where

#r# would be the distance from the center of mass to the end of the rod, or#L/2# .

#= m/L|[r^3/3]|_(-L/2)^(L/2)#

#= m/L[(L/2)^3/3 - (-L/2)^3/3]#

#= m/L[L^3/24 + L^3/24]#

#= m/L[L^3/12]#

#= color(green)(1/12mL^2)#

Lastly, we should recognize that the frequency is not equal to

#2pif = omega#

So, the **torque needed** should be:

#color(blue)(sum vectau) = vectau_(vec"F"_"app") = I_"cm"baralpha#

#= (mL^2)/12(vecomega_f)/t#

#= (("3 kg"cdot("4 m")^2)/12)((2pi*"2 s"^(-1))/("3 s"))#

#= color(blue)(5.bar"33"pi)# #color(blue)("N"cdot"m")#