Guest

The moment of inertia of cylinder about its own axis is equal to its moment of inertia about an Axis passing through its centre and normal to its length the ratio of length and radius

The moment of inertia of cylinder about its own axis is equal to its moment of inertia about an Axis passing through its centre and normal to its length the ratio of length and radius

Grade:11

5 Answers

Arun
25750 Points
4 years ago
Dear Sumit
 
 
L2 = 3R2
as MI of a solid cylinder about its own axis is ½ MR2
and MI about an axis passing through its centre of gravity and perpendicular to its length is
M(L2/12 + R2/4)
 ½ MR2 = M(L2/12 + R2/4)
R2/2 = L2/12 + R2/4
R2/2 – R2/4 = L2/12
L2 = 3R2
L1/2 = 31/2 R
Khimraj
3007 Points
4 years ago
½ MR2 = M(L2/12 + R2/4)
R2/2 = L2/12 + R2/4
R2/2 – R2/4 = L2/12
L2 = 3R2
..........................................................
aswanth nayak
100 Points
4 years ago
 
Dear Student
 
 
L2 = 3R2
as MI of a solid cylinder about its own axis is ½ MR2
and MI about an axis passing through its centre of gravity and perpendicular to its length is
M(L2/12 + R2/4)
 ½ MR2 = M(L2/12 + R2/4)
R2/2 = L2/12 + R2/4
R2/2 – R2/4 = L2/12
L2 = 3R2
L1/2 = 31/2 R
Yuvraj kamra
13 Points
4 years ago
as we know m.o.i of cyl. through central axis is mr^2/2.  and m.o.i of cyl. thr. axis passing thr. centre & normal to length is m(l^2/12 + r^2/4).  hence a.t.s. mr^2/2=m(l^2/12 + r^2/4). : r^2=l^2/6 + r^2/2 :r^2/2=l^2/6 :r=l/3^1/2
 
kapil chahar
18 Points
3 years ago
L2 = 3R2
as MI of a solid cylinder about its own axis is ½ MR2
and MI about an axis passing through its centre of gravity and perpendicular to its length is
M(L2/12 + R2/4)
 ½ MR2 = M(L2/12 + R2/4)
R2/2 = L2/12 + R2/4
R2/2 – R2/4 = L2/12
L2 = 3R2
L1/2 = 31/2 R

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free