Answer:
The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s
Explanation:
Moment of inertia is given as;
I = ¹/₁₂×ML² + 2mr²
where;
I is the moment of inertia
M is the mass of the rod = 0.19 kg
L is the length of the rod = 0.43 m
m is the mass of the bead = 0.038 kg
r is the distance of one bead
Initial moment of inertial is given as;
[tex]I_i = \frac{1}{12}ML^2 +2mr_1^2[/tex]
Final moment of inertia is also given as
[tex]I_f= \frac{1}{12}ML^2 +2mr_2^2[/tex]
Angular momentum is the product of angular speed and moment of inertia;
= Iω
From the principle of conservation of angular momentum;
[tex]I_i \omega_i = I__f } \omega_f[/tex]
[tex](\frac{1}{12}ML^2 +2mr_1^2) \omega_i = (\frac{1}{12}ML^2 +2mr_2^2) \omega_f[/tex]
Given;
ωi = 12 rad/s
r₁ = 10.0 cm = 0.1 m
r₂ = 10.0cm/4 = 2.5 cm = 0.025 m
Substitute these values in the above equation, we will have;
[tex](\frac{1}{12}*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = (\frac{1}{12}*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\\\\0.04425 =0.002975\ \omega_f\\\\\omega_f = \frac{0.04425}{0.002975} = 14.87\ rad/s[/tex]
Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s
