Question

A uniform rod of length $'6L'$ and mass $'8 \,m'$ is pivoted at its centre $'C'$. Two masses $'m'$ and $'2m'$ with speed $2v$, $v$ as shown strikes the rod and stick to the rod. Initially the rod is at rest. Due to impact, if it rotates with angular velocity $'\omega'$ then $'\omega'$ will be

• A. $\frac{v}{5L}$
• B. zero
• C. $\frac{8v}{6L}$
• D. $\frac{11v}{3L}$

Answer 1

Answer: (A)

A rod of length 6Land mass 8mis given in figure,

When two masses strike the rod, then angular momentum imparted to rod,
$L_{1}+L_{2}=2 m v(L)+m(2 v)(2 L)$
$=6 m v L$
Now, after striking of masses to rod the angular momentum of complete rod about the centre $C$,
$L_{\text {rod }} =k \omega$...(i)
where, $\omega=$ angular velocity of the rod and $I=$ moment of inertia of the rod.
The moment of inertia of rod
$I=\frac{M I^{2}}{12}=\frac{8 m(6 L)^{2}}{12}=24 m L^{2}$
$\therefore M = 8m $ and $I = 6L$(given)
Now, moment of inertia of two masses after the striking to $I$ rod
$I_1 = 2m(L)^2 = 2mL^2$
and $I_2 = m(2L)^2 = 4mL^2$
$\therefore $ The net moment of the inertia about the centre of rod,
$I=24 m L^{2}+2 m L^{2}+4 m L^{2}$
$\Rightarrow I =30 mL ^{2}$
By putting this value in Eq. (i), we get
$L_{\text {rod }}=30 M L^{2} \omega$
From the law of conservation of angular momentum,
$L_{1}+L_{2}=L_{\text {rod }}$
$\Rightarrow 6 m v L=30 m L^{2} \omega$
$\Rightarrow \omega=\frac{v}{5 L}$
Hence, the anguler velocity of the rod is $\frac{v}{5L}$