Question

Inititial angular velocity of a circular disc of mass M is 'omega 1'
.Then 2 small spheres of mass m are attached gently to diametrically opposite points on the edge of the disc . What is the final angular velocity of the disc?

Answer 1

Answer:

OMEGA2 IS YOUR ANSWER i.e (M/M+2m)omega1

Explanation:

CONSERVE THE ANGULAR MOMENTUM ABOUT DISC CENTRE

i.e (i1)OMEGA1=(i2)omega2 - LET THIS BE OUR EQ1

here i1 is MR^2 AND i2 is = MR^2 +mR^2+mR^2

HERE R IS RADIUS OF DISC WHICH WILL CANCEL OUT IN EQ1

Answer 2

initial angular velocity of a circular disc of mass M is w,then two small sphere of mass M is attached gently to two diametrically opposite points on the edge of the disc.

Intial moment of inertia of disc will be:

I₀ =( 1/2) MR²

Considering the spheres as a points of mass at a distance R form center,

Final Moment of inertia with two spheres attached diametrically will be:

I = (1/2) MR² + 2mR²

Since, there is no external torque applied on the disc, its angular momentum will be conserved.

i.e Intial angular momentum = Final angular momentum

Let final angular velocity of the disc be ω₂.

or, I₀ω₁ = Iω₂

or, ω₂ = I₀ω₁ / I

or, ω₂ = [ M / (M + 4m)] * ω₁

Therefore, Final angular velocity of the disc is  [ M / (M + 4m)] * ω₁.