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 normal to its length; then:

Question

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 normal to its length; then: (A) L=R (B) L=√3 R (C) L=R / √3 (D) L=0

Tardigrade Admin · Accepted Answer

Given I1=I2 ⇒ (1/2) M R2=(m L2/12)+(m R2/4) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2309cb48bbdfe67161f96d9d0142350b-.png" /> solve to get L=√3 R

Q. 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 normal to its length; then:

$L = R$

$L = 3 R$

$L = R / 3$

$L = 0$

Given $I_{1} = I_{2}$
$\Rightarrow \frac{1}{2} M R^{2} = \frac{m L ^{2}}{12} + \frac{m R ^{2}}{4}$

solve to get $L = 3 R$

Q. 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 normal to its length; then:

L=R

L=3​R

L=R/3​

L=0