The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio lR such that the moment of inertia is minimum ?
Answers
Answered by
0
Answer:
Given,
The length of the cylinder =l,
Radius of cylinder=R
And I=moment of inertia.
Now,
I=
4
mR
2
+
12
ml
2
I=
4
m
(R
2
+
3
l
2
)
I=
4
m
(
πl
V
+
3
l
2
) (∵V=πR
2
l)
Now differentiate Iwith respect to l
So,
dl
dI
=
4
m
(
πl
2
−V
+
3
2l
)
For maxima and minima,
dl
dI
=0
So,
4
m
(
πl
2
−V
+
3
2l
)=0
⇒
πl
2
V
=
3
2l
l
R
2
=
3
2l
(∵V=πR
2
l)
⇒
R
2
l
2
=
2
3
⇒
R
2
l
2
=
2
3
R
l
=
2
3
Similar questions