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A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 kg 0.5 m Find the moment of inertia of the wagon wheel for rotation about its axis. Answer in units of kg · m2 .

Answer :

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

[tex]I_1=m_1r^2[/tex]

[tex]I_1=7\times (0.25)^2=0.437\ kgm^2[/tex]

The moment of inertia of the rod about one end is given by :

[tex]I_2=\dfrac{m_2l^2}{3}[/tex]

l = r

[tex]I_2=\dfrac{m_2r^2}{3}[/tex]

[tex]I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2[/tex]

For 6 spokes, [tex]I_2=0.025\times 6=0.15\ kgm^2[/tex]

So, the net moment of inertia of the wagon is :

[tex]I=I_1+I_2[/tex]

[tex]I=0.437+0.15=0.587\ kgm^2[/tex]

So, the moment of inertia of the wagon wheel for rotation about its axis is [tex]0.587\ kgm^2[/tex]. Hence, this is the required solution.

onyebuchinnaji

The moment of inertia of the wagon about its axis is 0.588 kgm².

Moment of inertia of the wagon

The moment of inertia of the wagon about its axis is determined from the product of the mass and radius of the wheel.

I = mr²

where;

  • m is mass of the wagon
  • r is the radius

I = 7 x (0.25)²

I = 0.438 kgm²

Moment of inertia of the 6 spokes

I = 6(¹/₃ML²)

I = 6 x ¹/₃ x 1.2 x 0.25²

I = 0.15 kgm²

Total moment of inertia = 0.438 kgm² +  0.15 kgm² = 0.588 kgm².

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