The uniform solid block in the figure shown has mass 0.192 kg and edge lengths a

Question

The uniform solid block in the figure shown has mass 0.192 kg and edge lengths alpha = 3.95cm, b = 8.90 cm and C = 1.50 cm its rotational inertia about a parallel axix through the center of mass is I_com = (M/12)(a^2 + b^2) (a) Find the distance between the rotation axis and the center of mass of the block, h. (b) Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces, (as shown) (c) If the block is rotating about that point with an angular speed of 31.0 rotational kinetic energy of the block?

Explanation / Answer

a) h = sqrt[ (a/2)^2 + (b/2)^2 ] = sqrt[ (3.95/2)^2 + (8.90/2)^2]

h = 4.87 cm = 0.0487 m

b) I = Icom + m h^2

I = [ (0.192 ( 0.0395^2 + 0.089^2) / 12 ] + [ 0.192 x 0.0487^2 ]

I = 6.07 x 10^-4 kg m^2

c) KE = I w^2 / 2

= 6.07 x 10^-4 x 37^2 / 2 = 0.416 J

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