Answer:
A) ω₂ = 28 rev/min
B) ω_3 = 28 rev/min
Explanation:
A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).
Thus;
I₁ = ((1/12)ML²) + 2(mr²)
Where;
M is mass of rod = 0.03 kg
L is length of rod = 0.4m
m is mass of small rings = 0.02 kg
r is radius of small rings = 0.05 m
Thus;
I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)
I₁ = 0.0021 kg.m²
Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;
I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)
I₂ = 0.0036 kg.m²
From conservation of angular momentum, we know that:
I₁ω₁ = I₂ω₂
We are given ω₁ = 48 rev/min.
Thus; plugging in the relevant values;
0.0021 x 48 = 0.0036ω₂
0.1008 = 0.0036ω₂
ω₂ = 0.1008/0.0036
ω₂ = 28 rev/min
b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.
Thus,
I₂ω₂ = I_3•ω_3
So,0.0021 x 48 = 0.0036ω_3
ω_3 = 28 rev/min
