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The net work done in accelerating a propeller from rest to an angular speed of 124 rad/s is 2351.3 J. What is the moment of inertia of the propeller?

Answer in units of kg · m².

Answer :

The moment of inertia of the propeller will be "0.3058 k.gm²".

Moment of Inertia

According to the question,

Angular speed, ω = 124 rad/s

Net work done, W = 2351.3 J

By using Work-energy principle,

→ Net work done (W) = Kinetic energy (K.E)

We know,

→ K.E = [tex]\frac{1}{2}[/tex] × l × ω²

then,

→ W = [tex]\frac{1}{2}[/tex] × l × ω²

By substituting the values, we get

2351.3 = [tex]\frac{1}{2}[/tex] × l × (124)²

hence,

The moment of inertia,

→ I = 2351.3 × [tex]\frac{2}{(124)^2}[/tex]

= 0.3058 k.gm²

Thus the answer above is appropriate.

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Rewritten by : Barada

Answer:

0.3058 kgm²

Explanation:

Using the work-energy principle;

The net workdone(W) in accelerating the propeller is the increase in rotational kinetic energy (K.E) of the propeller. i.e;

W = K.E

But the rotational kinetic energy (K.E) of a body is half of the product of its rotational inertia (I) and the square of its angular velocity (ω) as follows;

K.E = [tex]\frac{1}{2}[/tex] x I x ω²

Recall that;

W = K.E

This means that;

W = [tex]\frac{1}{2}[/tex] x I x ω² -----------------(i)

Now, from the question;

I = rotational inertia or moment of inertia is unknown

ω = angular speed = 124rad/s

W = net work done = 2351.3J

Substituting the values of ω and W into equation (i) gives;

=> 2351.3 = [tex]\frac{1}{2}[/tex] x I x 124²

Making I the subject of the formula gives;

=> I = 2351.3 x 2 / 124²

=> I = 0.3058 kgm²

Therefore the moment of Inertia of the propeller is 0.3058 kgm²