Question

Figure is an overhead view of a thin uniform rod of length $\mathbf{0}\mathbf{.600}\mathbf{}\mathbf{m}$and mass Mrotating horizontally at $\mathbf{80.0}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$counter clock-wise about an axis through its centre. A particle of mass $\mathbf{M}\mathbf{/}\mathbf{3.00}$and travelling horizontally at speed $\mathbf{40.0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$hits the rod and sticks. The particle’s path is perpendicular to the rod at the instant of the hit, at a distance $\mathit{d}$from the rod’s centre.

(a) At what value of $\mathit{d}$are rod and particle stationary after the hit?

(b) In which direction do rod and particle rotate if $\mathit{d}$is greater than this value?

Short answer

1. Distance $d$ is$0.180\text{m.}$
2. Increasing the value of $d$would cause the rod to rotate in clockwise direction.

Step-by-step solution

Step 1: Given

$\begin{array}{l}\mathrm{L}=0.6\text{m}\\ \mathrm{\omega }=80\text{rad}/\text{sec}\\ \mathrm{v}=40\text{m/s}\end{array}$

Determining the concept

Use law of conservation of angular momentum to find the distance. Also, use the formula for angular momentum in terms of mass, velocity and distance to find the angular momentum.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

$\begin{array}{l}{\mathbf{L}}_{\mathbf{i}}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}\\ \mathbf{L}\mathbf{=}\mathbf{mvr}\end{array}$

Where,$\mathbf{L}$is angular momentum, $\mathbf{m}$is mass, $\mathbf{V}$is velocity and $\mathbf{r}$ is radius.

Determining the distance d

(a)

From the law of conservation of angular momentum,

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ -\mathrm{dmv}+\frac{1}{12}{\mathrm{ML}}^2\mathrm{\omega }=0\end{array}$

Negative sign is taken for the clockwise rotation.

So,

$\mathrm{d}=\frac{{\mathrm{ML}}^2\mathrm{\omega }}{12\mathrm{mv}}$

So,

$\begin{array}{l}\mathrm{d}=\frac{\mathrm{M}\times {0.6}^2\times 80}{12\times \frac{\mathrm{M}}{3}\times 40}\\ \mathrm{d}=0.18\text{0m}\end{array}$

Hence, the distance $\mathrm{d}$is$0.180\text{m.}$

Determining the effect of increase in  d

(b)

If the length$\mathrm{d}$ is increased, system rotates clockwise. This is because, when$d$ increases, term$mvd$in above equation would become more negative and hence would cause the clockwise rotation as per the sign conventions.

Hence, increasing the value of $d$would cause the rod to rotate in clockwise direction.

Therefore, using the law of conservation of angular momentum and angular momentum formula, the required distance and the sense of rotation can be found if the direction is changed.