First slide

Angular momentum

Question

A uniform bar of length 6 $α$ and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m moving in the same horizontal plane with speeds 2 v and v respectively strike the bar as shown in fig.
and stick to the bar after collision. Denoting angular velocity about the centre of mass, total energy and centre of mass velocity of $ω$ E and v_c respectively, we have after collision

Moderate

A

$v_{C} = 10 m / s e c$
B

$ω = \frac{3}{5} \frac{v}{a}$
C

$ω = \frac{V}{5 a}$
D

$E = {mv}^{2}$

Solution

Applying the law of conservation a of angular momentum, we have
$2 mva + m (2 v) (2 a)$
$= 2 {ma}^{2} ω + m (2 a)^{2} ω + 8 m \frac{(6 a)^{2}}{12} ω$ …………..(1)
Here $2 {ma}^{2}, m (2 a^{2}) and \{8 m (6 a^{2}) / 12\}$ are the moment of inertia of mass 2m, m and the bar respectively and $ω$ is the angular velocity bout the centre of mass. From eq. (1), we have
$\begin{array}{l} 6 mva = 2 {ma}^{2} ω + 4 {ma}^{2} ω + 24 {ma}^{2} ω \\ = 30 {ma}^{2} ω \end{array}$
$∴ ω = (\frac{V}{5 a})$ …………………………………(2)
(K'E') rotation
$\begin{array}{l} = \frac{1}{2} [(2 {ma}^{2}) ω^{2} + m (2 a)^{2} ω^{2} + 8 m \frac{(6 a)^{2}}{12} ω^{2}] \\ = \frac{1}{2} [30 {ma}^{2} ω^{2}] \\ = \frac{1}{2} [30 {ma}^{2} {(\frac{V}{5 a})}^{2}] = \frac{3}{5} {mv}^{2} \end{array}$
$∴ E = \frac{3}{5} {mv}^{2}$ ……………………………….(3)
The centre of mass does not charge in rotatory motion. The centre of mass is initially at rest. In the absence of external forces, the centre of mass continues in the same state of rest. Hence
V_C=0 …………………………….(4)

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