Question

A merry go round with a moment of inertia of 6.78x 10^3kg.m^2 is coasting at 2.20 rad/s.When a 75 kg man steps onto the rim the angular velocity decrease to 2.0 rad/s. The radius of the merry go round is?

Answer 1

Here angular momentum of the merry go is constant

Angular momentum is conserved

I₁ω₁=I₂ω₂

Given data

I₁=6.78x 10³kg.m²​​​​​​

ω₁=2. 2 rad/s

ω₂=2 rad /s

m =75 kg

Then

I₂=I₁ω₁/ω₂​​​​​​

I₂=(6.78x 10³kg.m²)(2.2)/(2)

I₂=7458 kg.m²​​​​​​

We know that When a 75 kg man steps onto the rim so

I₂=I₁+mr²​​​​​​

Then r =(( I₂-I₁​​​​​​) /m) ^1/2

Plugging the values

r=((7458-6780)/(75))^1/2​​​​​​

r = 3.007 meters

Radius of the merry go round r = 3.007 meters

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Answer 2

SOLUTION :

Let r meters be the radius of the merry-go-round.

Moment of inertia of the man weighing 75 kg

= 75 * r^2.       (Where r is the radius of the merry-go-round  ground)

M.I. of the merry-go-round is given = 6.78*10^3       kg-m^2

So, M.I. of after the man steps onto the merry-go-round

= 75 r^2 + 6.78*10^3     kg-m^2

Angular momentum remains unchanged .

So,

M.I. before l* angular speed before = M.I. afterwards * angular speed afterwards

=> 6.78*10^3 * 2.2 = (6.78*10^3 + 75 r^2) * 2.0

=> r^2 =( 6.78*10^3 * 2.2/2.0 - 6.78*10^3) / 75

=> r^2 = 9.04

=> r = 3.007 m = 3  m approximately (ANSWER)