Question

A uniform rod of length ${}^{\prime }6L^{\prime }$ and mass ${}^{\prime }8m^{\prime }$ is pivoted at its centre ${}^{\prime }{C}^{\prime }$. Two masses ${}^{\prime }{m}^{\prime }$ and ${}^{\prime }2m^{\prime }$ 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 ${}^{\prime }{\omega }^{\prime }$ then ${}^{\prime }{\omega }^{\prime }$ 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,
$L_1+L_2=2mv(L)+m(2v)(2L)$
$=6mvL$
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
$I=\frac{M{I}^2}{12}=\frac{8m(6L{)}^2}{12}=24m{L}^2$
$\therefore M=8m$ and $I=6L$(given)
Now, moment of inertia of two masses after the striking to $I$ rod
$I_1=2m(L{)}^2=2m{L}^2$
and $I_2=m(2L{)}^2=4m{L}^2$
$\therefore$ The net moment of the inertia about the centre of rod,
$I=24m{L}^2+2m{L}^2+4m{L}^2$
$\Rightarrow I=30m{L}^2$
By putting this value in Eq. (i), we get
$L_{\text{rod }}=30M{L}^2\omega$
From the law of conservation of angular momentum,
$L_1+L_2=L_{\text{rod }}$
$\Rightarrow 6mvL=30m{L}^2\omega$
$\Rightarrow \omega =\frac{v}{5L}$
Hence, the anguler velocity of the rod is $\frac{v}{5L}$