JEE Advanced 2016 · Paper 2 · Physics Q7

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PhysicsMultiple correct+4 / -2Question 7 of 54

Two thin circular discs of mass $m$ and $4m$ , having radii of $a$ and $2a$ , respectively, are rigidly fixed by a massless, rigid rod of length $l=\sqrt{24}\,a$ through their centers. 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 $\omega$ . The angular momentum of the entire assembly about the point 'O' is $\vec L$ (see the figure). Which of the following statement(s) is(are) true?

A small disc (mass m, radius a) and a large disc (mass 4m, radius 2a) joined through their centres by a rigid rod, both standing on a flat hatched surface. The rod's axis, extended (dashed), meets the surface at point O, through which a vertical z-axis passes. Distances l are marked from O to the small disc's centre and from the small disc's centre to the large disc's centre; the assembly spins with angular speed ω about the rod.

The center of mass of the assembly rotates about the

z

-axis with an angular speed of

\omega/5

The magnitude of angular momentum of center of mass of the assembly about the point O is

81ma^2\omega

The magnitude of angular momentum of the assembly about its center of mass is

17ma^2\omega/2

The magnitude of the

z

-component of

\vec L

55ma^2\omega

Answer and worked solution

Official answer: A

Precession. The assembly rolls like a cone with apex at O, where the line of the rod meets the surface. The instantaneous axis is the line of contact through O, so the total angular velocity has no vertical part. The precession about $z$ must cancel the vertical part of the spin:
$\Omega=\omega\sin\phi,$
where $\phi$ is the rod's inclination. Taking O as 5a from the small disc's centre (height $a$ ), $\sin\phi=\frac15$ and $\Omega=\omega/5$ . (A) is true.

About the centre of mass. The spin angular momentum is
$\left(\tfrac12ma^2+\tfrac12(4m)(2a)^2\right)\omega=\frac{17}2ma^2\omega,$
as in (C) (ignoring the small contribution of the precession).

The other options. (B) and (D) do not hold once the cosine of the inclination is included.

Why the key accepts two answers. The figure's lengths are not consistent: a rod of $\sqrt{24}\,a$ does not match the heights $a$ and $2a$ with O at 5a. The key therefore accepts (A) or (A), (C).

Answer: (A) (the key accepts A, or A and C)

Topics:Angular momentum and its conservation Rolling motion Centre of mass

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JEE Advanced 2016 · Paper 2, Physics question 7