find an expression for a moment of inertia of a thin rod rotating about an axis passing through its center perpendicular to it's length. plz help :/

the Axis Perpendicular to the Length and Passing through the Center. Take a uniform thin rod AB whose mass is M and length is l. YY’ axis passes through the center of the rod and perpendicular to the length of the rod. We have to calculate the moment of inertia about this axis. YY’.

Suppose, small piece dx is at a distance x from the YY' axis. Hence, mass of length dx is =(

l

M

)dx The moment of inertia of small piece about the YY' axis

dl=[(

l

M

)dx]x

2

Therefore, the moment of inertia of the complete rod about the YY' axis.

b) Moment of Inertia about the axis perpendicular to the the length and passing through the Corner If I

CD

is the moment of inertia about the axis CD perpendicular to the length of a thin rod and passing through the point A then, by theorem of parallel axis;

l

CD

=l

YY

′

+Md

2

=

12

Ml

2

+M(

2

l

)

2

l

CD

=

3

Ml

2

(2)

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

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explanation: