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`        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`
5 months ago

```							Dear Sumit  L2 = 3R2as MI of a solid cylinder about its own axis is ½ MR2and MI about an axis passing through its centre of gravity and perpendicular to its length isM(L2/12 + R2/4) ½ MR2 = M(L2/12 + R2/4)R2/2 = L2/12 + R2/4R2/2 – R2/4 = L2/12L2 = 3R2L1/2 = 31/2 R
```
5 months ago
```							½ MR2 = M(L2/12 + R2/4)R2/2 = L2/12 + R2/4R2/2 – R2/4 = L2/12L2 = 3R2..........................................................
```
5 months ago
```							 Dear Student  L2 = 3R2as MI of a solid cylinder about its own axis is ½ MR2and MI about an axis passing through its centre of gravity and perpendicular to its length isM(L2/12 + R2/4) ½ MR2 = M(L2/12 + R2/4)R2/2 = L2/12 + R2/4R2/2 – R2/4 = L2/12L2 = 3R2L1/2 = 31/2 R
```
5 months 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
```
4 months ago
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