Two particles each with mass m= 0.85 kg, are fastened to each other?

Question

, and
to a rotation axis at O, by two thin rods, each with length d= 5.6 cm and
mass M= 1.2 kg. The combination rotates around the rotation axis with
angular speed= 0.30 rad/s. Measured about O, what are the combination's: (A) rotational inertia
(B) kinetic energy

Answer 1

Hi!
(A)
By definition di=dm * X^2 , where di is the amount added to I when the mass dm is added to the system. I~Inertia . In this case for the particle X=d.

Each mass m=0.85 kg add to the system (delta I)1 = m * d^2.

Each rod has an Inertia of (delta I)2 = (1/3) M * d^2 . This is because for a rod you have to calculate:
S { dm * x^2 = (M/d)*dx * x^2 } ==> (M/d) * (1/3)(X^3) with X=d. => (1/3)M * d^2

so the total inertia of the system is I= 2*[ m*d^2+ (1/3) M*d^2] because we have 2 rods and tho particles.:
I= 7.84 * 10^(-3) kg / m^2. Where ^ is used for power!

(B) kinetic energy= (1/2) Inertia * (angularspeed)^2 =3.528 /10^4

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