Answer:
a) Angular velocity decreases
b) for L / 3. the frequency changes by 1 /√3
for R/√3 the frequency changes by √ 9
Explanation:
As in the problem they indicate that we have a torsion constant, we have a torsion pendulum, with angular velocity
w =√ I / k
The angular momentum of a cylinder with an axis of rotation passing through its center is
I = ½ m R²
We replace
w = √ ½ m R² / k
Angular velocity and frequency are related.
w = 2π f
f = w/2π
Let's analyze what happens with the requested changes. We will use the subscript zero for the initial dimensions. The dimensions have been reduced to 1/3 of the original values, but maintaining the wheel density
ρ = m₀ / V₀
With the new values
ρ = (m₀ / 3) / (V₀ / 3)
So that the density does not change, let's look for what change we must make in the volume of the cylinder
V = π r² L
We can make two changes
1) L = L₀/3, maintaining the same radius
The angular velocity is left with these changes
w’= √ ½ m₀/3 R₀² / k = 1 /√3 √ ½ m₀ R₀ / k
w’= w₀ /√3
Therefore the angular velocity decreases
Frequency is
f’/ 2pi = 1 /√3 f₀ / 2pi
f' = 1 /√3 f₀
2) R =√3 R₀, maintaining the same length
Angular velocity
w’= √ ½ m₀ / 3 (R₀² / 3) / k
w = 1 /√9 √ ½ m₀R₀² / k
w ’= 1 /√9 w₀
Angular velocity decreases
The angular frequency is
f ’= f₀ /√ 9
The frequency also decreases
