a thin rod of length L is suspended from one end and rotated with n rotations per second the rotational kinetic energy of rod will be ?
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Answered by
65
angular velocity w=2πn
Rotational ke=½iw²
where i=moment of inertia of rod = ml²/3
now rotational k.e=½(ml²/3)*(2πn)²
=2/3*mπ²n²l²
sorry i didn't see the term rotational..
it seems that only linear ke kinetic energy is to find
Rotational ke=½iw²
where i=moment of inertia of rod = ml²/3
now rotational k.e=½(ml²/3)*(2πn)²
=2/3*mπ²n²l²
sorry i didn't see the term rotational..
it seems that only linear ke kinetic energy is to find
RajveerShekhawat1:
no
y..
whatever enrgy is here that is rotational
don't.. make anyone. fool
it's not correct answer
I.. know
Answered by
17
w=2*pi*n
I=(ml^2)÷3
kE =1/2(iw^2). (1)put the value of w and I in eqn 1
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