QJEEAdv16Ph19cc: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph19cc Solution: Two thin circular discs of mass m and 4m, having radii of a and …

The following figure depicts the rolling condition of a circular disc:



For the smaller disc, we have the velocity expressed as

v = a $\omega$

For the bigger disc, we have the velocity expressed as

v = (2a) $\omega$ = 2a $\omega$

Now, let us consider that both discs roll and also rotate about point O.



$\bullet$ The angular momentum of a rotating body is given by

$\overrightarrow L = \overrightarrow r \times \overrightarrow p$

About point O, the combined angular momentum of the discs is given by

L = lma $\omega$ + 2l(4m)(2a $\omega$ )

where lma $\omega$ is due to the smaller disc and 2l(4m) (2a $\omega$ ) is due to the bigger disc.

Now, substituting the values in Eq. (1), we get

$L = m{a^2}\omega [\sqrt {24} + 16\sqrt {24} ]$

$= m{a^2}\omega (34\sqrt 6 ) \approx 83m{a^2}\omega$

Hence, option (B) is incorrect.

$\bullet$ Now, let us consider that z-component of $\overrightarrow L$ :

$\overrightarrow L = L\cos \theta$

$= 34\sqrt 6 m{a^2}\omega \left[ {{1 \over {\sqrt {{l^2} + {a^2}} }}} \right] = 34\sqrt 6 m{a^2}\omega \left( {{{\sqrt {24} } \over 6}} \right)$

= 81 ma² $\omega$

Hence, option (D) is incorrect.

$\bullet$ The centre of mass of the assembly about its centre of mass [i.e. the axis lavelled as (I) in the figure shown here] is

${L_{(I)}} = {{m{a^2}} \over 2}\omega + {{4m{{(2a)}^2}} \over 2}\omega = {{17m{a^2}} \over 2}\omega$

Hence, option (C) is correct.


$\bullet$ About z-axis, the centre of mass of the assembly rotates with an angular speed

${\omega _z} = \omega \sin \theta = \omega \left[ {{a \over {\sqrt {{l^2} + {a^2}} }}} \right] = {\omega \over 5}$

Hence, option (A) is correct.