Two thin discs each of mass 4.0 kg and radius 0.4 kg and radius 0.4 m are attached as shown in figure to form a rigidbody. What is the rotational inertia of this body about an axis perpendicular to the plane of disc $B$ and passing through its center?

1.6kg - m²

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3.2kg - m²

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6.4kg - m²

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0.8kg - m²

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Solution

The correct option is B
3.2kg - m²

$I_{B}$ = $I$ $o f s p h e r e$ $B$ $a b o u t c e n t r e o f m a s s +$ $I$ $o f s p h e r e$ $A$ $a b o u t c e n t r e o f m a s s + m a s s o f s p h e r e$ $A$ $\times$ $(d i s t a n c e b e t w e e n c e n t r e s o f A a n d B)^{2}$

= $\frac{m R^{2}}{2} + \frac{m R^{2}}{2} + m (2 R)^{2}$ = $5 m R^{2}$ = $5 \times 4 \times (0.4)^{2}$ = $3.2 k g / m^{2}$

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Two thin discs each of mass 4.0 kg and radius 0.4 m are attached as shown in figure to form a rigid body. What is the rotational inertia of this body about an axis perpendicular to the plane of disc $B$ and passing through its center?

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Two thin discs each of mass 4.0 kg and radius 0.4 kg and radius 0.4 m are attached as shown in figure to form a rigidbody. What is the rotational inertia of this body about an axis perpendicular to the plane of disc B and passing through its center?

The correct option is B 3.2kg - m2 IB = I of sphere B about centre of mass + I of sphere A about centre of mass + mass of sphere A × (distance between centres of Aand B)2 = mR22 + mR22 + m(2R)2 = 5mR2 = 5×4×(0.4)2 = 3.2 kg/m2

Two thin discs each of mass 4.0 kg and radius 0.4 kg and radius 0.4 m are attached as shown in figure to form a rigidbody. What is the rotational inertia of this body about an axis perpendicular to the plane of disc $B$ and passing through its center?