Question

The wheels of a wagon can be approximated as the combination of a thin outer hoop of radius h=0.156 m and mass 4.89 kg, and two thin crossed rods of mass 7.80 kg each. Imagine replacing the wagon wheels with uniform disks that are d=5.25 cm thick, made out of a material with a density of 7370 kg/m3. 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?

rd=? m

Answer 1

The radius of the disk of the new wheels should be 0.142 m. It was calculated from the moment of inertia of the old wheels, which is a combination of a thin outer hoop (radius = 0.156 m and mass = 4.89 kg) and two thin crossed rods (mass = 7.80 kg each one), and from the density (7370 kg/m³) and thickness (5.25 cm) of the disk.    

The moment of inertia of the old wheel ($I_{o}$) is given by the sum of the moment of inertia of the outer hoop and the two rods, as follows:

$I_{o} = I_{h} + 2*I_{r}$

$I_{o} = m_{h}r_{h}^{2} + 2*\frac{1}{12}m_{r}L_{r}^{2}$  (1)

Where:

$I_{h}$: is the moment of inertia of the outer hoop  

$I_{r}$: is the moment of inertia of the rods

$m_{h}$: is the mass of the hoop = 4.89 kg

$r_{h}$: is the radius of the hoop = 0.156 m

$m_{r}$: is the mass of each rod = 7.80 kg

$L_{r}$: is the length of each rods = $2*r_{h}$ = 2*0.156 m = 0.312 m  

Then, the moment of inertia of the old wheel is (eq 1):

$I_{o} = 4.89 kg*(0.156 m)^{2} + 2*\frac{1}{12}*7.80 kg*(0.312 m)^{2} = 0.246 kg*m^{2}$

The moment of inertia of the new wheel ($I_{n}$) is the same as the old wheel, so:

$I_{n} = I_{o}$

$\frac{1}{2}m_{d}r_{d}^{2} = 0.246 kg*m^{2}$   (2)

The mass of the disk, can be calcualted from its density

$m_{d} = d*V_{d}$ (3)

The volume of the disk is equal to a cylinder's volume:

$V_{d} = \pi r_{d}^{2} h$   (4)

Where h is the thickness of the disk = 0.0525 m

By entering equations (3) and (4) into (2), we have:

$\frac{1}{2}d*V_{d}*r_{d}^{2} = 0.246 kg*m^{2}$  

$\frac{1}{2}d(\pi r_{d}^{2}*h)r_{d}^{2} = 0.246 kg*m^{2}$  

$\frac{1}{2}d*\pi*h*r_{d}^{4} = 0.246 kg*m^{2}$    

$r_{d} = (\frac{2*0.246 kg*m^{2}}{\pi*d*h})^{1/4} = (\frac{2*0.246 kg*m^{2}}{\pi*7370 kg/m^{3}*0.0525 m})^{1/4} = 0.142 m$

Therefore, the radius of the disk should be 0.142 m.

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