Question

two point masses m and 2m are joined by light rod of length d.the moment of inertia of the system about an axis passing through its centre of mass and making an angle 30° with the light rod will be​

Answer 1

Point masses m and 2m are joined by light rod of length d . Axis passes through thr centre of mass and making angle 30° with light rod.

To find :

Moment of Inertia of system

Calculation:

First we need to find out the Centre of Mass .

Distance of Centre of mass from mass m will be :

$\bar x = \dfrac{(2m)d}{2m + m} = \dfrac{2d}{3}$

Similarly Centre of Mass from 2m will be :

$\bar {x2} = \dfrac{(m)d}{2m + m} = \dfrac{d}{3}$

Now perpendicular distance of 2m from axis will be :

$(r2) \perp = \dfrac{d}{3} \times \sin( 30\degree) = \dfrac{d}{6}$

Again, perpendicular distance of m from axis will be :

$(r1) \perp = \dfrac{2d}{3} \times \sin( 30\degree) = \dfrac{d}{3}$

So moment of Inertia of system will be :

$\boxed{ \red{MI = 2m {( \dfrac{d}{6}) }^{2} + m {( \dfrac{d}{3} )}^{2} = \dfrac{m {d}^{2} }{6} }}$

Answer 2

Explanation:

$\bf \huge \: Question$

• two point masses m and 2m are joined by light rod of length d.the moment of inertia of the system about an axis passing through its centre of mass and making an angle 30° with the light rod will be

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$\bf \huge \: </strong><strong>Give</strong><strong>n$

• Two point masses m and 2m
• Light rod of length d.
• The moment of inertia of the system about an axis passing through its centre of mass
• making an angle 30°

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$\bf \huge \: </strong><strong>To</strong><strong>\: </strong><strong>Find</strong><strong> </strong><strong>$

• The light rod will be

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___________________________$\bf \huge \: </strong><strong>s</strong><strong>o</strong><strong>l</strong><strong>u</strong><strong>t</strong><strong>i</strong><strong>o</strong><strong>n</strong><strong>\: </strong><strong>$

We Find the Centre of Mass .

We know

Distance of Centre of mass from mass m will be :-

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Then Accourding to the question :-

putting the value:-

$\bf \bar x = \dfrac{(2m)d}{2m + m} = \dfrac{2d}{3}$

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Then ,

We know that

Given to the Question

2m are joined

$\bf \bar {x2} = \dfrac{(m)d}{2m + m} = \dfrac{d}{3}$

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Given to the Question :-

two point masses m

Putting the value

$\bf (r2) \perp = \dfrac{d}{3} \times \sin( 30\degree) = \dfrac{d}{6}$

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We Find out perpendicular distance :-

$\bf(r1) \perp = \dfrac{2d}{3} \times \sin( 30\degree) = \dfrac{d}{3}$

We get the moment of Inertia of system :-

$\bf MI = 2m {( \dfrac{d}{6}) }^{2} + m {( \dfrac{d}{3} )}^{2} = \dfrac{m {d}^{2} }{6}$

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HENCE the light rod will be $</strong><strong>\bf</strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong>MI</strong><strong> = 2m {( \dfrac{d}{6}) }^{2} + m {( \dfrac{d}{3} )}^{2} = \dfrac{m {d}^{2} }{6}$

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