Question

A thin ring has mass m=4 kg and radius r = 1 m. Find its moment of inertia about an axis passing through its centre which makes an angle of exactly pi/4 radians

Answer 1

Answer:

(1/2)MR^2

Explanation:

moment of inertia of the circular ring for the given condition is this

Answer 2

The moment of intertia about an axis passing through its center is 2 kg m².

Explanation:

Given that,

Mass of ring =4 kg

Rdius = 1 m

Angle $\theta=\dfrac{\pi}{4}$

We need to calculate the moment of intertia about an axis passing through its center

Using formula of moment of inertia

$I=MR^2$

The moment of inertia does not depend on angle.

By perpendicular axis theorem

$I_{x}+I_{y}=I_{z}$

Here, $I_{x}=I_{y}$

$2I_{x}=I_{z}=MR^2$

Put the value of moment of inertia

$I=\dfrac{MR^2}{2}$

Put the value into the formula

$I=\dfrac{4\times1}{2}$

$I=2\ kg m^2$

Hence, The moment of intertia about an axis passing through its center is 2 kg m².

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Topic : moment of inertia

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