A uniform cylinder has a radius r and length l. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length, then
Answers
Answer: l = r√3
Explanation:
The radius of the cylinder is given as = r
The length of the cylinder is given as = l
Let the mass of the cylinder be denoted as “m”.
We know,
The moment of inertia of the cylinder about an axis passing through its centre and normal to its circular face,
I₁ = [mr²]/2 …… (i)
And,
The moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to the length of the cylinder,
I₂ = [(ml²)/12 + (mr²)/4] ….. (ii)
We are given that the moment of inertia of the cylinder in both the cases is equal to each other, therefore, on equating the equations (i) & (ii), we get
I₁ = I₂
⇒ [mr²]/2 = [(ml²)/12 + (mr²)/4]
⇒ [mr²]/2 – [mr²]/4 = [ml²]/12
⇒ [1/2 – 1/4]*[mr²] = [ml²]/12
⇒ [¼]*[mr²] = [1/12]*[ml²]
cancelling the similar terms
⇒ r² = 1/3 * l²
⇒ l² = 3* r²
taking square roots on both sides
⇒ l = r√3
Thus, If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length, then the length of the cylinder becomes square root 3 times the radius of the cylinder.