A uniform, solid bar with mass m and length L rotates with angular velocity omega about an axle on one end of the bar. What is the bar’s kinetic energy? Hint: you need an integral.

Answer 1

We know that rotational kinetic energy (E_K-r) is defined by:

E_K-r = 1/2Iω² where I is the rotational moment of inertia and ω is the angular velocity.

We also know that ω is constant and we know that I = mr² where r is the distance away from the center of rotation. But the mass changes as you move out away from the center of rotation in this case, hence the integral.

Yo do not specify what the cross section of the bar is (circular like a rod or is it square or something else?). Because of that, I'll just assume it has a cross sectional area of "A". Then a sketch might look like this:

Since the mass is "M" and the volume is A•L then the density (ρ) is M/(A•L) and the mass of any random cross section at some random location "r" from the axis of rotation is ρ•A•dr. We can replace ρ with M/(A•L) to get the mass of the cross section(M_CS) is [M/(A•L)]•A•dr or M_CS = M/Ldr. therefore, I_CS = (M/Ldr)r²

To find "I" of the whole bar, we integrate: I = ₀∫^LM/Lr² dr = (M/L)• ₀∫^L r² dr

Hopefully you can integrate this function then plug what you get as "I" into E_K-r = 1/2Iω²