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Grade: 11
        
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
3 months ago

Answers : (4)

Arun
22770 Points
							
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
3 months ago
Khimraj
3008 Points
							
½ MR2 = M(L2/12 + R2/4)
R2/2 = L2/12 + R2/4
R2/2 – R2/4 = L2/12
L2 = 3R2
..........................................................
3 months ago
aswanth nayak
66 Points
							
 
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
2 months ago
Yuvraj kamra
13 Points
							
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
 
2 months ago
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