For the case of the lefthand
(righthand) example from the previous
pages, the proof is not complicated.
Since,
r1
(r2)
is perpendicular
to F, one has for the modulus of the torque
τ
τ = r1 F
(
r2 F
)
Let us define for the acceleration
of ω
α = dω/dt
Then, since the object has no initial angular velocity,
we get for the angular velocity at instant
t the expression
ω(t)
= αt
Furthermore, since the initial angle of rotation of the object equals zero,
we find
φ(t)
= αt2 / 2
The force F stops to acting on the system
for φ=2π, say at the instant
τ. Then
2π = φ(τ)
= ατ2 / 2
→
τ = √ ( 4π/α )
So, we obtain for the final angular velocity squared
ω2(τ)
= ( ατ )2
= 4π α
For the final kinetic energy we have then
2π Iα = I ω2(τ) /2
= Ekin
= 2π r1 F
(
2π r2 F
)
= 2π τ
Hence,
I dω/dt
= Iα = τ
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