Respuesta :
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
The conservation of angular momentum allows finding the results for the angular velocity of the system are:
a) With the rings at the ends of the rod. w = 12 rpm
b) When the rings leave the rod. w = 15 rpm.
Given parameters.
- Rod mass M = 0.0300 kg
- Length L = 0.400 m.
- Ring mass m = 0.0200 kg.
- Distance to the rod scepter r = 0.0500 m.
- Angular speed w₀ = 48 rev / min.
To find.
a) Angular velocity with the rings at the ends.
b) Angular velocity without rings.
The angular momentum is equivalent to the linear momentum for rotation, it is a vector magnitude, which is conserved if the system is isolated.
L = I w
where L is the angular momentum, I the moment of inertia and w the angular velocity.
The symmetrical bodies moment of inertia is tabulated.
- Rod with axis per cu center I₁ = [tex]\frac{1}{12}[/tex] M L²
- Approximate the rings to point masses I₂ = m L²
a) We define the system to be formed by the rod and the rings, therefore the system is isolated and the angular momentum is conserved.
Initial instant. Before releasing the rings
L₀ = (I₁ + 2 I₂₀) w₀
The two comes from having two rings.
Final instant. When the rings reach the end of the rod.
[tex]L_f = ( I_1 + 2 I_{2f} ) w[/tex]
Angular momentum is conserved.
[tex]L_o = L_f \\(I_1 + 2 I_2_o) w_o = ( I_1 + 2 I_{2f}) w[/tex]
[tex]w = \frac{I_1 + 2I_2_o}{I_1 + 2 I_2_f} \ w_o[/tex]
We calculate
I₁ = [tex]\frac{1}{12}[/tex] 0.03 0.4² = 4 10-4 kg m²
I₂₀ = 0.02 0.05² = 5 10-5 kg m²
[tex]I_2_f[/tex] = 0.02 0.2² = 8 10-4 kg m²
We substitute
[tex]w = \frac{4 \ 10^{-4} + 2 \ 5 \ 10^{-5}}{4\ 10^{-4} + 2 \ 8 \ 10^{-4}} \ 48[/tex]
w = 12 rpm
b) when the rings have left the rod
Initial instant. With the rings at the ends of the rod.
L₀ = (I₁ +2 [tex]I_{2f}[/tex]) [tex]w_f[/tex]
Final moment. Without the rings
[tex]L_f = I_1 w[/tex]
[tex]L_o = L_f[/tex]
[tex]w = \frac{I_1 + 2 \ I_2_f}{I_1} \ w_f[/tex]
Let's calculate.
[tex]w = \frac{4 \ 10 ^{-4} + 2 \ 8 10^{-5}}{4 \ 10^{-4}} \ 12[/tex]
w = 15 rpm
In conclusion using the conservation of angular momentum we can find the results for the angular velocity of the system are:
a) With the rings at the ends of the rod, w = 12 rpm
b) When the rings leave the rod, w = 15 rpm.
Learn more about angular momentum here: brainly.com/question/25677703
