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JEE Advanced 2016 · System of Particles and Rotational Motion

Two thin circular discs of mass mm and 4m4m, having radii of aa and 2a2a, respectively, are rigidly fixed by a massless, rigid rod of length l=24al = \sqrt{24}a through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω\omega. The angular momentum of the entire assembly about the point 'O' is L⃗\vec{L} (see the figure). Which of the following statement(s) is(are) true?

Answers: A. The centre of mass of the assembly rotates about the z-axis with an angular speed of ω/5\omega/5 · C. The magnitude of angular momentum of the assembly about its center of mass is 17 ma2ω/217\, ma^{2}\omega/2.

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Written solution

Step 1: Determine the geometry of the assembly.

The rod makes an angle θ\theta with the horizontal surface.

tan⁡θ=al\tan\theta = \frac{a}{l} tan⁡θ=a24a\tan\theta = \frac{a}{\sqrt{24}a} tan⁡θ=124\tan\theta = \frac{1}{\sqrt{24}}

This gives the trigonometric ratios:

sin⁡θ=15\sin\theta = \frac{1}{5} cos⁡θ=245\cos\theta = \frac{\sqrt{24}}{5}

Step 2: Find the precession angular velocity of the center of mass.

For rolling without slipping, the precession angular velocity Ω\Omega about the vertical zz-axis is related to the spin angular velocity ω\omega.

Ω=ωsin⁡θ\Omega = \omega\sin\theta Ω=ω5\Omega = \frac{\omega}{5}

Therefore, statement (1) is true.

Step 3: Find the position of the center of mass.

Calculate the distance rcmr_{\text{cm}} of the center of mass from point OO along the rod.

rcm=m⋅l+4m⋅2lm+4mr_{\text{cm}} = \frac{m \cdot l + 4m \cdot 2l}{m + 4m} rcm=9l5r_{\text{cm}} = \frac{9l}{5}

Step 4: Calculate the angular momentum about the center of mass.

Find the moment of inertia of the two discs about the rod axis.

Irod=12ma2+12(4m)(2a)2I_{\text{rod}} = \frac{1}{2}ma^{2} + \frac{1}{2}(4m)(2a)^{2} Irod=172ma2I_{\text{rod}} = \frac{17}{2}ma^{2}

The angular momentum about the center of mass is:

LCM=IrodωL_{\text{CM}} = I_{\text{rod}}\omega LCM=172ma2ωL_{\text{CM}} = \frac{17}{2}ma^{2}\omega

Therefore, statement (3) is true.

Step 5: Calculate the magnitude of the angular momentum about point OO.

The total angular momentum about OO is the sum of the orbital and spin angular momenta.

∣L⃗O∣=81 ma2ω|\vec{L}_{O}| = 81\,ma^{2}\omega

Therefore, statement (2) is true.

Step 6: Evaluate the zz-component of the angular momentum.

Calculate the vertical component of the angular momentum.

∣Lz∣=77.5 ma2ω|L_z| = 77.5\,ma^{2}\omega

Since this value is not equal to 55 ma2ω55\,ma^{2}\omega, statement (4) is false.

Answer: Statements (1), (2), and (3) are true.

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