Answer:
8.59 rad/s
Explanation:
Given that:
A thin uniform cylindrical turntable radius = 2.7 m
with a mass (M) = 22 kg
initial angular speed (ω₁) = 12 rad/s
Mass of the clump of the clay (m) = 11 kg
Diameter (d) from the point of rotation = 1.7 m
We are to find the final angular velocity (ω₂) ,To do that; we apply the conservation of annular momentum; which is as follows:
L₁ = L₂
l₁ω₁ = l₂ω₂
( 0.5 × M × r²) × ω₁ = (0.5 × M × r² + md²) ω₂
Making ω₂ the subject of the formula ; we have:
[tex]\omega_2 = \frac{(0.5*M*r^2)* \omega_1}{(0.5*M*r^2+md^2)}[/tex]
[tex]\omega_2 = \frac{(0.5*22*2.7^2)* 12}{(0.5*22*2.7^2+11*1.7^2)}[/tex]
[tex]\omega_2 = 8.58 rad/s[/tex]
Hence, the angular speed of the clay and turntable = 8.59 rad/s
