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Nataliya [291]
3 years ago
15

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 k

g 0.5 m Find the moment of inertia of the wagon wheel for rotation about its axis. Answer in units of kg · m2 .
Physics
1 answer:
Arte-miy333 [17]3 years ago
4 0

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 :

I_1=m_1r^2

I_1=7\times (0.25)^2=0.437\ kgm^2

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

I_2=\dfrac{m_2l^2}{3}

l = r

I_2=\dfrac{m_2r^2}{3}

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

For 6 spokes, I_2=0.025\times 6=0.15\ kgm^2

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

I=I_1+I_2

I=0.437+0.15=0.587\ kgm^2

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

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<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
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Answer:

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