Question

A uniform ring of mass 10 kg and diameter 0.40m rotaes with a uniform speed of 2100 rotations per min. Find the moment of inertia and angular momentum of the ring about its geometric axis​

Answer 1

hi mate

here is your answer

Given, diameter= 0.4m

So,radius = 04/2 = 0.2m

Mass of ring = 10kg

Now,

1)Angular momentum

=>>L = Iω

and ω = 2πν.

(ν = 2100rpm = 2100/60 r/s)

Moment of inertia ,

I = MR²

=>I = 10 × (0.2)² = 10 × 0.04 = 0.4 Kg.m²

So,L = 0.4 × 2π × 2100/60

=> L= 87.92Kgm²/s.

2)moment of inertia

we know,

I = MR²

=>I = 10 × (0.2)² = 10 × 0.04 = 0.4 Kg.m²

=>I = 0.4 Kg.m²

hope it helps

BE BRAINLY///

Answer 2

Answer:

this is the answer best answer