Two particles, each with mass m = 4.4 g, are fastened to each other and to a rot

Question

Two particles, each with mass m = 4.4 g, are fastened to each other and to a rotation axis at P, by two thin rods, each with length L = 0.86 m and a mass of 7.7 g, as shown. The combination rotates around the rotation axis with an angular velocity of 14.4 rad/s.

A) Find the rotational inertia of the combination about P.

B) What is the kinetic energy associated with the rotation about P?

I have the answer to (A), I only need the correct answer for (B).

Two particles, each with mass m = 4.4 g, are fastened to each other and to a rotation axis at P, by two thin rods, each with length L = 0.86 m and a mass of 7.7 g, as shown. The combination rotates around the rotation axis with an angular velocity of 14.4 rad/s. A) Find the rotational inertia of the combination about P. B) What is the kinetic energy associated with the rotation about P? I have the answer to (A), I only need the correct answer for (B).

Explanation / Answer

a)

As mass of each rod is m=7.7 g and their combined length is 2L=2*0.86 =1.72 m ,so moment of inertia of the rods w.r.t axis at point p is

Irod=(1/3)(2M)(2L)2 =(1/3)(2*7.7*10-3)(2*0.86)2 =0.0152 Kg-m2

Moment of inertia of 2 balls about P is

Iball=mL2+m(2L)2=5mL2 =5*(4.4*10-3)*0.862

Iball=0.01627 Kg-m2

So Net rotational inertia about point P is

I=0.01627+0.0152

I=0.0315 Kg-m2 or 3.15*10-2 Kg-m2

b)

Rotational KE

KErot=(1/2)IW2 =(1/2)*0.0315*14.42

KErot=3.26 J

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