Question

A Texas cockroach of mass $\mathbf{0}\mathbf{.17}\mathbf{}\mathbf{kg}\mathbf{}$ runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius 15 cm, rotational inertia $\mathbf{5}\mathbf{.0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{kg}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$, and frictionless bearings. The cockroach’s speed (relative to the ground) is $\mathbf{2.0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, and the lazy Susan turns clockwise with angular speed ${\mathbf{\omega }}_{\mathbf{0}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.8}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}\mathbf{.}$The cockroach finds a bread crumb on the rim and, of course, stops.

(a) What is the angular speed of the lazy Susan after the cockroach stops?

(b) Is mechanical energy conserved as it stops?

Short answer

1. The angular speed of the lazy Susan after the cockroach does stop is, ${\mathrm{\omega }}_{\mathrm{f}}=4.2\mathrm{rad}/\mathrm{s}$
2. The mechanical energy is not conserved as the cockroach stops.

Step-by-step solution

Given

1. The mass of the cockroach,$\text{M}=0.17\text{kg}.$
2. The rim of the lazy Susan (a circular disk mounted on a vertical axle) is,$\text{R}=15\text{cm}=0.15\text{m}$
3. The rotation inertia of the lazy Susan (a circular disk mounted on a vertical axle) is,$\text{I}=5.0\times {10}^{-3}{\text{kgm}}^2$.
4. The speed of the cockroach relative to the ground is,$\text{v}=2.0\text{m}/\text{s}$
5. The angular speed of the Susan is,${\text{ω}}_0=2.8\text{rad}/\text{s}$.

To understand the concept

Using the law of conservation of angular momentum, find the angular speed of the lazy Susan after the cockroach does stop.

Formula:

The law conservation of angular momentum is,

Initial angular momentum of the system = Final angular momentum of the system

Calculate the angular speed of the lazy Susan after the cockroach stops 

(a)

According to the conservation of the angular momentum,

Initial angular momentum of the system = Final angular momentum of the system

$\begin{array}{c}\Rightarrow \text{mvR}-{\text{Iω}}_0=-({\text{mR}}^2+\text{I}){\text{ω}}_{\text{f}}\\ \Rightarrow {\text{ω}}_{\text{f}}=\frac{\text{mvR}-{\text{Iω}}_0}{({\text{mR}}^2+\text{I})}\end{array}$ (The clockwise motion is taken as negative.)

Therefore, the angular speed of the lazy Susan after the cockroach does stop is, ${\text{ω}}_{\text{f}}=4.2\mathrm{rad}/\mathrm{s}$.

Calculate the mechanical energy conserved

(b)

According to the law of conservation of energy,

The Initial mechanical energy of the cockroach-Susan system is,

$\begin{array}{c}{\text{K}}_{\text{i}}=\text{K}.{\text{E}}_{\text{iCockroach}}+\text{K}.{\text{E}}_{\text{iSusan}}\\ =\frac{1}{2}{\text{mv}}^2+\frac{1}{2}{\text{Iω}}_0^2\\ {\text{K}}_{\text{i}}=\frac{1}{2}\left(0.17\right){\left(2\right)}^2+\frac{1}{2}(5\times {10}^{-3}){\left(2.8\right)}^2\\ {\text{K}}_{\text{i}}=0.34\text{J}\end{array}$

The final mechanical energy of the cockroach-Susan system

$\begin{array}{c}{\text{K}}_{\text{f}}=\text{K}.{\text{E}}_{\text{fCockroach}}+\text{K}.{\text{E}}_{\text{fSusan}}\\ \Rightarrow {\text{K}}_{\text{f}}=0+\frac{1}{2}{\text{Iω}}_{\text{f}}^2\\ \Rightarrow {\text{K}}_{\text{f}}=\frac{1}{2}(5\times {10}^{-3}){\left(4.2\right)}^2\\ \Rightarrow {\text{K}}_{\text{f}}=19.6\times {10}^{-3}=44.1\times {10}^{-3}\text{J}\end{array}$

As,${\text{K}}_{\text{i}}\ne {\text{K}}_{\text{f}},$ the mechanical energy of the cockroach-Susan system is not conserved