Question 612: Centre of Mass and Rotational Motion

Question

A uniform rod of length ' 6 L ' and mass ' 8 m ' is pivoted at its centre ' $C$ '. Two masses ' $m$ ' and ' $2 m^{\prime}$ with speed $2 v, v$ 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

Accepted Answer

A rod of length 6L and mass 8 m is given in figure,
When two masses strike the rod, then angular momentum imparted to rod,
$$\begin{aligned}
L_1+L_2 & =2 m v(L)+m(2 v)(2 L) \
& =6 m v L
\end{aligned}$$
Now, after striking of masses to rod the angular momentum of complete rod about the centre $C$,
$$L_{	ext {od }}=10\quad 	ext{... (i)}$$
where, $\omega=$ angular velocity of the rod
and $\quad I=$ moment of inertia of the rod.
The moment of inertia of rod
$$I=\frac{M I^2}{12}=\frac{8 m(6 L)^2}{12}=24 m L^2$$
$	herefore M=8 \mathrm{~m}$ and $I=6 L$ (given)
Now, moment of inertia of two masses after the striking to $I$ rod
$$I_1=2 m(L)^2=2 m L^2$$
and $$I_2=m(2 L)^2=4 m L^2$$
$	herefore$ The net moment of the inertia about the centre of rod,
$$\begin{aligned}
& I=24 m L^2+2 m L^2+4 m L^2 \
\Rightarrow \quad & I=30 \mathrm{~mL}^2
\end{aligned}$$
By putting this value in Eq. (i), we get
$$L_{	ext {rod }}=30 M L^2 \omega$$
From the law of conservation of angular momentum,
$$\begin{aligned}
& & L_1+L_2 & =L_{	ext {rod }} \
\Rightarrow & & 6 m v L & =30 m L^2 \omega \
\Rightarrow & & \omega & =\frac{v}{5 L}
\end{aligned}$$
Hence, the anguler velocity of the rod is $\frac{v}{5 L}$.

Answer

$\frac{V}{5 L}$

Answer

zero

Answer

$\frac{8 v}{6 L}$

Answer

$\frac{11 v}{3 L}$

Centre of Mass and Rotational Motion

A uniform rod of length ' 6 L ' and mass ' 8 m ' is pivoted at its centre ' C '. Two masses ' m ' and ' 2 m^{\prime} with speed 2 v, 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

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