A solid cylinder of radius $\dfrac{R}{3}$ and length $\dfrac{L}{2}$ is removed along the central axis. Find ratio of initial moment of inertia and moment of inertia of removed cylinder.

Question

A solid sphere of mass $5$ kg and radius $10$ cm is kept in contact with another solid sphere of mass $10$ kg and radius $20$ cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ___________ kg$\cdot$m$^2$.

Answer 1

Step 1: Moment of inertia of the original cylinder

For a solid cylinder of mass M and radius R, the moment of inertia about its central axis is:

I_initial = (1/2) M R²

Step 2: Determine the mass of the removed cylinder

Radius of removed cylinder = R/3
Length of removed cylinder = L/2

Volume of removed cylinder:

V_removed = π (R/3)² (L/2)

V_removed = πR²L / 18

Volume of original cylinder:

V_original = πR²L

Since mass is proportional to volume:

M_removed = M × (V_removed / V_original) = M/18

Step 3: Moment of inertia of the removed cylinder

Moment of inertia of a solid cylinder about its own axis is:

I = (1/2) m r²

Thus,

I_removed = (1/2) × (M/18) × (R/3)²

I_removed = (1/2) × (M/18) × (R²/9)

I_removed = MR² / 324

Step 4: Calculate the required ratio

Ratio = I_initial / I_removed

= [(1/2)MR²] / [MR²/324]

= (1/2) × 324

= 162

Final Answer:

The ratio of the initial moment of inertia to the moment of inertia of the removed cylinder is
162

A solid cylinder of radius $\dfrac{R}{3}$ and length $\dfrac{L}{2}$ is removed along the central axis. Find ratio of initial moment of inertia and moment of inertia of removed cylinder.

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The Correct Option is A

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