(D) Let the mass of each sphere be $m = 5 \ kg$ and the length of the rod be $L = 1 \ m$.Case $1$: Axis passing through $A$ and perpendicular to the r

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(D) Let the mass of each sphere be $m = 5 \ kg$ and the length of the rod be $L = 1 \ m$.Case $1$: Axis passing through $A$ and perpendicular to the rod.The moment of inertia $I_1$ is given by:$I_1 = m_A \cdot (0)^2 + m_B \cdot (L)^2 = 0 + m \cdot (1)^2 = m$Case $2$: Axis passing through the center of the rod and perpendicular to it.The moment of inertia $I_2$ is given by:$I_2 = m_A \cdot (L/2)^2 + m_B \cdot (L/2)^2 = m \cdot (0.5)^2 + m \cdot (0.5)^2 = 0.25m + 0.25m = 0.5m = m/2$Ratio of moments of inertia:$I_1 / I_2 = m / (m/2) = 2 / 1$Thus,the ratio is $2 : 1$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia and Radius of gyration

Medium

Two spheres $A$ and $B$,each of mass $5 \ kg$,are attached to the ends of a light rod of length $1 \ m$. Treating the spheres as point masses,find the ratio of the moment of inertia of the system about an axis passing through $A$ to that about an axis passing through the center of the rod,both axes being perpendicular to the rod.

A
$1 : 1$
B
$1 : 4$
C
$1 : 2$
D
$2 : 1$

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Chapter

System of Particles and Rotational Motion

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Moment of Inertia and Radius of gyration

Two discs of the same material and thickness have radii $0.2\, m$ and $0.6\, m.$ Their moments of inertia about their axes will be in the ratio

Medium

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Two identical rods each of mass $M$ and length $l$ are joined in a crossed position as shown in the figure. The moment of inertia of this system about a bisector ($B_1$ or $B_2$) is:

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Match the following columns ($R=$ radius,$k=$ radius of gyration):
Column $I$ Column $II$
$(A)$ 'k' for a solid sphere rotating about its tangent $(P)$ $\sqrt{2}R$
$(B)$ 'k' for a ring rotating about its tangent perpendicular to its plane $(Q)$ $\frac{R}{2}$
$(C)$ 'k' for a uniform solid right circular cone rotating about its central axis $(R)$ $\frac{\sqrt{7}}{\sqrt{5}}R$
$(D)$ 'k' for a uniform disc rotating about its diameter $(S)$ $\frac{\sqrt{3}}{\sqrt{10}}R$

Column $I$	Column $II$
$(A)$ 'k' for a solid sphere rotating about its tangent	$(P)$ $\sqrt{2}R$
$(B)$ 'k' for a ring rotating about its tangent perpendicular to its plane	$(Q)$ $\frac{R}{2}$
$(C)$ 'k' for a uniform solid right circular cone rotating about its central axis	$(R)$ $\frac{\sqrt{7}}{\sqrt{5}}R$
$(D)$ 'k' for a uniform disc rotating about its diameter	$(S)$ $\frac{\sqrt{3}}{\sqrt{10}}R$

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If the mass and radius of a hollow sphere are $M$ and $R$ respectively,then its moment of inertia with respect to its diameter is:

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Two rods of equal length $(l)$ and equal mass $M$ are kept along the $x$ and $y$ axes respectively,such that their center of mass lies at the origin. The moment of inertia about the line $y = x$ is ...........

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Two discs of the same material and thickness have radii $0.2\, m$ and $0.6\, m.$ Their moments of inertia about their axes will be in the ratio

Two identical rods each of mass $M$ and length $l$ are joined in a crossed position as shown in the figure. The moment of inertia of this system about a bisector ($B_1$ or $B_2$) is:

If the mass and radius of a hollow sphere are $M$ and $R$ respectively,then its moment of inertia with respect to its diameter is:

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