O 4/5/2016 11:00 PM A 25/100 f 4/5/2016 12:27 PM Gradebook Print Calculator Peri

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O 4/5/2016 11:00 PM A 25/100 f 4/5/2016 12:27 PM Gradebook Print Calculator Periodic Table Question 4 of 16 ncorrect Map A pling The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius n F0.527 m and mass 5.46 kg, and two thin crossed rods of mass 7.37 kg each. You would like to replace the wheels with uniform disks that are 0.0525 m thick, made out of a material with a density of 8750 kilograms per cubic meter. If the new wheel is to have the same moment of inertia about its center as the old wheel about its center, what should the radius of the disk be? Number 0.0525 m. A Previous ® Give Up & View solution Check Answer Next Ext O Hint

Explanation / Answer

The rotational inertia of the hoop is

I = MR2

= (5.46 kg)(0.527m)2

= 1.51 64 kg m2

The rotational inertia of a rod through its center is

I_rod (1/12)ML2

= (1/12)(7.37 kg)(1.054 m)2

= 0.682 kg m^2

total inertia

I_total = 1.51 64 kg m2+ 2( 0.682) = 2.88 kg m^2

let t = thickness of disk, the volume of the disk is r2t and its mass is r2t where is the density.

Since the rotational inertia of a disk is (1/2)MR2

(1/2)(r2t)(r2) = Itot   or (1/2)r4t = Itot.

r = (2Itot/t)1/4

= ((2 * 2.88 kg m2)/( * 8750 kg/m3 * 0.0525 m))1/4

= 0.251 m

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