Grade 11Mechanics

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

Sumit Anand

6 Years agoGrade 11

5 Answers

Arun

6 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

L^1/2= 3^1/2 R

Khimraj

6 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

6 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

L^1/2= 3^1/2 R

Yuvraj kamra

6 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

5 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

L^1/2= 3^1/2 R

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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

Sumit Anand

5 Answers

Arun

Khimraj

aswanth nayak

Yuvraj kamra

kapil chahar

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