Figure is an overhead view of a thin uniform rod of length $\mathbf{0}\mathbf{.800}\mathbf{}\mathbf{m}$ and mass $\mathbf{M}$rotating horizontally at angular speed $\mathbf{20.0}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$about an axis through its centre. A particle of mass $\mathbf{M}\mathbf{/}\mathbf{3.00}$initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle’s speed role="math" localid="1660999204837" ${\mathit{v}}_{\mathbf{p}}$is$\mathbf{6.00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}$ greater than the speed of the rod end just after ejection, what is the value of ${\mathbf{v}}_{\mathbf{p}}$?

1. Length of the rod,$L=0.8\text{m}$
2. Mass of the rod$\mathrm{M}$,
3. Angular velocity of the rod,$\mathrm{\omega }=18\text{rad}/\text{s}$
4. Mass of the particle, $\mathrm{m}=\frac{\mathrm{M}}{3}$

Using the law of conservation of momentum, find the final speed of the particle. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

1. Angular momentum of the rod, $\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}{\mathbf{ML}}^{\mathbf{2}}$
2. Initial angular momentum = Final angular momentum

Where,Iis moment of inertia, M is mass and L is length of a rod.

According to law of conservation of angular momentum:

Total angular momentum before ejection = Total angular momentum after ejection

${\mathrm{L}}_1={\mathrm{L}}_2$

Momentum of the particle before the ejection,

$\begin{array}{l}{\mathrm{L}}_1=\mathrm{I\omega }\\ {\mathrm{L}}_1=\mathrm{m}{\left(\frac{\mathrm{L}}{2}\right)}^2\mathrm{\omega }\end{array}$

Momentum of the rod,

$\begin{array}{l}{\mathrm{L}}_2=\mathrm{I\omega }\\ {\mathrm{L}}_2=\frac{1}{12}{\mathrm{ML}}^2\mathrm{\omega }\end{array}$

Momentum of the particle after the ejection,

$\begin{array}{l}\mathrm{L}{\text{'}}_1=\mathrm{I\omega }\\ \mathrm{L}{\text{'}}_1=\mathrm{m}{\left(\frac{\mathrm{L}}{2}\right)}^2\left(\frac{{\mathrm{v}}_{\mathrm{p}}}{\left(\frac{\mathrm{L}}{2}\right)}\right)\\ \mathrm{L}{\text{'}}_1=\mathrm{m}\left(\frac{\mathrm{L}}{2}\right){\mathrm{v}}_{\mathrm{p}}\end{array}$

Momentum of the rod,

$\begin{array}{l}\mathrm{L}{\text{'}}_2=\mathrm{I\omega }\text{'}\\ \mathrm{L}{\text{'}}_2=\frac{1}{12}{\mathrm{ML}}^2\mathrm{\omega }\text{'}\end{array}$

Total momentum would be sum of the momentum of the rod and particle,

$\mathrm{L}{\text{'}}_1+{\mathrm{L}}_2^{\text{'}}={\mathrm{L}}_1+{\mathrm{L}}_2$

Putting the values,

${\mathrm{mv}}_{\mathrm{p}}\left(\frac{\mathrm{L}}{2}\right)+\left(\frac{1}{12}\right){\mathrm{ML}}^2{\mathrm{\omega }}^{\text{'}}=\mathrm{m}{\left(\frac{\mathrm{L}}{2}\right)}^2\mathrm{\omega }+\left(\frac{1}{12}\right){\mathrm{ML}}^2\mathrm{\omega }$

Using the given information, angular velocity of the rod can be written as,

${\mathrm{\omega }}^{\text{'}}=\frac{{\mathrm{v}}_{\mathrm{p}}-6}{\frac{0.8}{2}}$

${\mathrm{mv}}_{\mathrm{p}}\left(\frac{0.8}{2}\right)+\left(\frac{1}{12}\right)3\mathrm{m}\times 0{.8}^2\times \frac{{\mathrm{v}}_{\mathrm{p}}-6}{\frac{0.8}{2}}=\mathrm{m}{\left(\frac{0.8}{2}\right)}^2\times 18+\left(\frac{1}{12}\right)3\mathrm{m}\times 0{.8}^2\times 18$

By solving above equation for $v_p,$

${\mathrm{v}}_{\mathrm{p}}=11\text{m}/\text{s}$

Hence, the particle’s speed is, ${\mathrm{v}}_{\mathrm{p}}=11\text{m}/\text{s}$.