The wheels of a wagon can be approximated as the combination of a thin outer hoo
ID: 1482909 • Letter: T
Question
The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius rh = 0.209 m and mass 4.51 kg, and two thin crossed rods of mass 7.80 kg each. You would like to replace the wheels with uniform disks that are 0.0588 m thick, made out of a material with a density of 5990 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?
the response below is from anonymous, and is incorrect.
Expressing what we know in equation form: (1) Id must = Io, where: Id is the moment of inertia (MoI) of the new disk; and Io = MoI of the old wheel (2) Io = Ihoop + 2 * Irod (3) Ihoop = Mhoop * Rhoop^2 = 4.51 * 0.209^2 = 0.197 kg·m^2 (4) Irod = Mrod * Lrod^2 / 12 = 7.80 * (2 * Rhoop)^2 / 12 = 7.80 * (2 * 0.209)^2 / 12 = 0.1136 kg·m^2 Substituting the values into (2) we get: (5) Io = 0.197 + 2 * 0.1136 = 0.4242 kg·m^2 The MoI for a disk is: (6) Id = m * r^2 / 2 We'll need to calculate the mass from its volume V and density : (7) m = V * The volume is area X thickness: (8) V = * r^2 * 0.0588 Substituting into (7): (9) m = ( * r^2 * 0.0588) * and substituting for m into (6): (10) Id = (( * r^2 * 0.0588) * ) * r^2 / 2 = * r^4 * 0.0588 * / 2 Solving for r the radius: (11) r^4 = Id / ( * 0.0588 * / 2) Since Id must equal Io, which we have from (5), (12) r = ( Io / ( * 0.0588 * / 2) )^(1/4) = ( 0.4242 / (3.14 * 0.0525 * 5990 / 2) )^(1/4) =
0.1712 m = radius of the disk <----- THIS IS INCORRECT ACCORDING TO SAPLING LEARNING
Explanation / Answer
Given that
The wheels of a wagon can be approximated as the combination of a thin outer hoop, of radius(rh) = 0.209 m
The wheel of mass(m) = 4.51 kg
The two thin crossed rods of mass (M) =7.80 kg each.
You would like to replace the wheels with uniform disks that are (t) = 0.0588 m thick
The material with a density of (p) = 5990 kilograms per cubic meter
The rotational inertia of the hoop is mrh2 =(4.51kg)(0.209m)2 =0.197kg.m2
The rotational inertia of the rod through the center is =(1/12)ML2 = (1/12)*(7.80kg)(2*0.209m)2 =0.1135kg.m2
There are two thin crossed rod then total inertia becomes =2*0.1135kg.m2=0.227kg.m2
Now the total rotational inertia of the wheel is
Itot =0.197kg.m2+0.227kg.m2=0.4240kg.m2
Now mass of the second wheel is (m) =Volume *density
=pir2h*5990 kg/m3
=3.14*(r)2( 0.0588 m)*5990 kg/m3
=1105.945*r2kg
Now the moment of inertia of the disk is (I) =mr2/2 =(1105.945*r2kg)*r2/2
552.972r4kg.m2 =0.4240kg.m2
r4 =0.4240/552.972 =7.667*10-4
then r =0.7667m or 76.67cm
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