Find the kinetic energy of the bar

Question

A 50 cm long metal bar with mass 25 grams is attached to a pivot with a massless string 80 cm long. Find the kinetic energy of the bar when it is rotated at a rate of 4 revolutions per second.

Answer 1

The problem does not mention anything about the mass density of the rod, so

I assume that it is uniform.

Let

b = 50cm be the length of the bar,

s = 80cm be the length of the string,

m = 25g be the mass of the bar,

ρ = m/b be the linear mass density of the bar,

ω = 8π s^-1 be the angular velocity.

Consider a small section of the bar of length dr, some distance r from the pivot.

That section will have mass ρdr and velocity rω.

Therefore its kinetic energy will be (1/2)(rω)²ρdr.

The kinetic energy of the whole rod will be the sum over all those small sections,

i.e., the definite integral from r = s to r = s+b.

Let's first do the indefinite integral

∫(1/2)(rω)²ρdr =

(1/2)ω²ρ ∫r²dr =

(1/2)ω²ρ (1/3)r³ =

(1/6)ω²ρr³

So the definite integral is

(1/6)ω²ρ(s+b)³ - (1/6)ω²ρs³ =

(1/6)ω²ρ((s+b)³ - s³)

Substituting actual numbers, the kinetic energy is

(1/6)×64×π²×(25/50)(130³ - 80³) = 88694845 g cm²/s² = 8.87 J