Question

What is the moment of inertia I of an object that rotates at 13.0 rev/min13.0 rev/min about an axis and has a rotational kinetic energy of 16.0 J?

Answer 1

Answer:

The moment of inertia of the object is 17.276 kilogram-square meters.

Explanation:

According to the statement, we find that object has rotation and no translation. From Rotation Physics we get that rotational kinetic energy ($K_{R}$), measured in joules, is represented by the following formula:

$K_{R} = \frac{1}{2}\cdot I_{G}\cdot \omega^{2}$ (Eq. 1)

Where:

$I_{G}$ - Moment of inertia with respect to center of mass, measured in kilogram-square meters.

$\omega$ - Angular speed, measured in radians per second.

Now we clear the moment of inertia:

$I_{G} = \frac{2\cdot K_{R}}{\omega^{2}}$

If we know that $K_{R} = 16\,J$ and $\omega \approx 1.361\,\frac{rad}{s}$, then the moment of inertia of the object is:

$I_{G} = \frac{2\cdot (16\,J)}{\left(1.361\,\frac{rad}{s} \right)^{2}}$

$I_{G} =17.276\,kg\cdot m^{2}$

The moment of inertia of the object is 17.276 kilogram-square meters.

Answer 2

The moment of inertia of the object will be "17.276 kg/m²".

Moment of inertia

Rotational Kinetic energy, $K_R$ = 16 J

Angular speed, ω = 1.361 rad/s

By using the Rotation Physics, the relation will be:

→ $K_R$ = $\frac{1}{2}$ × $I_G$ × ω²

the,

The moment of inertia be:

→ $I_G$ = $\frac{2\times K_R}{\omega^2}$

By substituting the values, we get

       = $\frac{2\times 16}{(1.361)^2}$

       = $\frac{32}{(1.361)^2}$

       = 17.276 kg.m²

Thus the above answer is correct.  

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