A230 g, 50.0-cm-diameter turntable rotates on frictionless bearings at 63.0 rpm. a 21.0 g block sits at the center of the turntable. a compressed spring shoots the block radically outward along a frictionless groove in the surface of the turntable.
what is the turntable's rotation angular velocity when the block reaches the outer edge?

ω₂ = 5.578 rads⁻¹

Explanation:

The moment of inertia of,

turn table =

block =

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

( + )×6.597 = ( + )×ω₂

ω₂ = 5.578 rads⁻¹

3.2x10^7

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s