Answer:
ω₂ = 5.578 rads⁻¹
Explanation:
The moment of inertia of,
turn table = [tex]\frac{1}{2}mr^{2}[/tex]
block = [tex]mr^{2}[/tex]
By the law of conservation of angular momentum, as no external torque acts on system the total angular momentum of the system when block sits at center and when block at the outer edge must be same.
Also to find ω, ω = 2πf where f is rpm
= 2π×63/60 = 6.597
So
I₁ω₁ = I₂ω₂
([tex]\frac{1}{2}0.23(0.5)^{2}[/tex] + [tex]0.021(0)^{2}[/tex])×6.597 = ([tex]\frac{1}{2}0.23(0.5)^{2}[/tex] + [tex]0.021(0.5)^{2}[/tex])×ω₂
ω₂ = 5.578 rads⁻¹
This really isn't an answer, but a correction to the tutors answers. I can't comment on her post for some reason, so I figured I would say it down here. You have to divide the diameter by 2 to get the radius. She forgot to do that and used 0.5cm^2 when it should have been 0.25cm^2. Besides that, her answer will give you the correct thing.
