The value entered above (.189) is not correct. Please help! The hints it gave ar

Question

The value entered above (.189) is not correct. Please help! The hints it gave are listed below.

The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius = 0.580 m and mass 5.65 kg, and two thin crossed rods of mass 9.52 kg each. You would like to replace the wheels with uniform disks that are 0.0462 m thick, made out of a material with a density of 8290 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 .189 im 9 0.0462 m

Explanation / Answer

let the radius of the new wheel is r ,

Now,

momemt of inertia , I = moment of inertia of hoop + moment of inertia of rods

I = 5.65 * 0.58^2 + 2 * 9.52 * (2 * 0.580)^2/12

I= 4.04 Kg.m^2

Now, for the disk to have same moment of inertia ,

I = 0.5 * m * r^2

4.04 = 0.5 * 8290 * pi*r^2 * 0.0462 * r^2

solving for r

r = 0.286 m

the radius of the disk wheel is 0.286 m

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