Discussion Forum

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 assemply 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 assemply about the point 'O' is L (see the figure). Which of the following statement(s) is (are) true?

Answer:

Explanation:

$\cos\theta =\frac{1}{\sqrt{l^{2}+a^{2}}}=\frac{\sqrt{24}}{5}$

(c) Velcocity of point P:aω=1Ω then

Ω = $\frac{a\omega}{1}=$ Angular velocity of C M w.r.t

Point O. Angular velocity fo CM w.r.t Z - axis = Ω cosθ

$\omega_{CM-Z}=\frac{a\omega}{1}\frac{\sqrt{24}}{5}=\frac{a\omega}{\sqrt{24}a}\frac{\sqrt{24}}{5}$

$\omega_{CM-Z}=\frac{a\omega}{5}$

(a) $L_{z}=L_{CM}-0^{\cos\theta}-L_{D}-_{CM}sin\theta$

$=\frac{81\sqrt{24}}{5}a^{2}m\omega \times\frac{\sqrt{24}}{5}-\frac{17ma^{2}\omega}{2}\times\frac{1}{\sqrt{24}}$

$=\frac{81\times 24 ma^{2}\omega}{25}-\frac{17ma^{2}\omega}{2\sqrt{24}}$

(b) $L_{CM-O}=(5m)[\frac{9l}{5}Ω]\frac{9l}{5}=\frac{81ml^{2}\omega}{5}$

  $=\frac{81ml^{2}}{5}\times\frac{a\omega}{l}$

$L_{CM-O}=\frac{81 mla\omega}{5}=\frac{81\sqrt{24}a^{2}m\omega}{5}$

(d)  $L_{D-CM}=\frac{ma^{2}}{2}\omega+\frac{4m(2a)^{2}}{2}\omega=\frac{17ma^{2}\omega}{2}$