Question

Calculate the M.I of a circular wheel of mass 1.4kg and radius 15cm about an axis passing through

its centre and normal to its plane, assuming entire mass of the wheel is concentrated at its rim​

Answer 1

Concept:

• Moment of inertia is defined as the measure of effort required to rotate the distributed mass.
• That distributed mass is assumed to mostly as on centre of mass.

★ But, here in the question,

• The mass in concentrated at its rim, instead at its centre.

Solution:

so , we can make the system SIMPLE, by considering it as particle system

• where we have to find the inertia of the distributed mass which is concentrated at distance 'r' , just like an point mass of mass 'm'

See the attached figure for understanding.

_________________

so, M.I for the axis passing for the 'm'

$\implies \bf I = mR^2$

$\longrightarrow \rm I = 1.4(0.15)^2$

$\longrightarrow \rm I = 1.4\times 0.15\times 0.15$

$\longrightarrow \bf I = 3.15 \times 10^{-2}\:kgm^2$

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Answer 2

Explanation:

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}$

$\large\bf{\underline{\red{VERIFIED✔}}}$

Concept:

• Moment of inertia is defined as the measure of effort required to rotate the distributed mass.
• That distributed mass is assumed to mostly as on centre of mass.

★ But, here in the question,

The mass in concentrated at its rim, instead at its centre.

Solution:

so , we can make the system SIMPLE, by considering it as particle system

where we have to find the inertia of the distributed mass which is concentrated at distance 'r' , just like an point mass of mass 'm'

See the attached figure for understanding.

_________________

so, M.I for the axis passing for the 'm'

$\implies \bf I = mR^2$

$\longrightarrow \rm I = 1.4(0.15)^2$

$\longrightarrow \rm I = 1.4\times 0.15\times 0.15$

$\longrightarrow \bf I = 3.15 \times 10^{-2}\:kgm^2$

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