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Lunna [17]
3 years ago
15

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

tia of the particle described in the problem introduction with respect to the axis about which it is rotating.
Physics
1 answer:
stira [4]3 years ago
3 0

Answer:

See explanation

Explanation:

We have a mass m revolving around an axis with an angular speed \omega, the distance from the axis is r. We are given:

\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]

and also the formula which states that the kinetic rotational energy of a body is:

K =\frac{1}{2}I\omega^2.

Now we use the kinetic energy formula

K =\frac{1}{2}mv^2

where v is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

v=\omega r

After replacing in the previous equation we get:

K =\frac{1}{2}m(\omega r)^2

now we have the following:

K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2

therefore:

mr^2=I

then the moment of inertia will be:

I = 13*(0.5)^2=3.25 [Kg*m^2]

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