In the figure, two particles, each with mass m = 0.83 kg, are fastened to each o

Question

In the figure, two particles, each with mass m = 0.83 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.4 cm and mass M = 1.0 kg. The combination rotates around the rotation axis with angular speed ? = 0.27 rad/s. Measured about O, what is the combination's (a) rotational inertia and (b) kinetic energy?

In the figure, two particles, each with mass m = 0.83 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 5.4 cm and mass M = 1.0 kg. The combination rotates around the rotation axis with angular speed ? = 0.27 rad/s. Measured about O, what is the combination's (a) rotational inertia and (b) kinetic energy?

Explanation / Answer

(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]

I = 2 [ (0.83 kg ) (5.4*10^-2 m )^2 + ( 1/3) (1.0 kg ) (5.4*10^ -2 m )^2 ]
I= 6.78* 10^(-3) kg / m^2.

(B)

kinetic energy= (1/2) Inertia * (angularspeed)^2

= ( 0.5 )(0.00678 kg m^2 )(0.27 rad/s)^2

= 0.00024 J

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