Question

A thin rod of mass $ 0.9 kg$ and length $ 1 m$ is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass $ 0.1 kg$ moving in a straight line with a velocity of $ 80 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ hits the rod at its bottommost point and sticks to it (see figure). The angular speed $ \left(in \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)$ of the rod immediately after the collision will be __________.

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Answer 1

Step1: Given data and assumptions.

Mass of rod, $M=0.9kg$

Suspended length of the rod, $L=1m$

Mass of particles, $m=0.1kg$

The initial velocity of the particle, $u=80\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step2: Find the angular speed of the rod immediately after the collision.

We know that,

According to the law of conservation of angular momentum about the pivotal point.

$L_i=L_f$

Where $L_i$ is the initial momentum and $L_f$ is the final momentum.

Also,

$\Rightarrow$$muL+I_{rod}{\omega }_i=\left(I_{particle}+I_{rod}\right){\omega }_{final}$

Where $I$ is the inertia and $\omega$ is the angular speed.

$\Rightarrow$ $muL+0=\left(I_{particle}+I_{rod}\right){\omega }_{final}$ $\left[\because I_{rod}{\omega }_i=0\right]$

$\Rightarrow$ $muL=\left(m{R}^2+\frac{M{R}^2}{3}\right){\omega }_{final}$ $\left[\because I_{particle}=mRand{I}_{rod}=\frac{M{R}^2}{3}\right]$

Where $R$ is the radius.

$\Rightarrow$$0.1\times 80\times 1=\left(0.1\times 1^2+\frac{0.9\times 1^2}{3}\right){\omega }_{final}$ $\left[\because R=L=1m\right]$

$\Rightarrow$ $8=\frac{4}{10}{\omega }_{final}$

$\Rightarrow$ ${\omega }_{final}=\frac{80}{4}$

$\Rightarrow$ ${\omega }_{final}=20\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Hence, the angular speed (in rad/s) of the rod immediately after the collision will be $20\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.