Question

A man stands on a platform that is rotating (without friction) with an angular speed of $\mathbf{1}\mathbf{.2}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{s}\mathbf{;}$his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is localid="1660979279335" $\mathbf{6}\mathbf{.0}\mathbf{}\mathbf{kg}\mathbf{.}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{.}$If by moving the bricks the man decreases the rotational inertia of the system to 2.0 kg.m².

(a) What are the resulting angular speed of the platform?

(b) What is the ratio of the new kinetic energy of the system to the original kinetic energy?

(c) What source provided the added kinetic energy?

Short answer

1. The angular speed of the platform after the brick is moved closer to the body of man is$3.6\text{rev}/\text{s}$.
2. The ratio of the new kinetic energy to the initial$\mathrm{K}.\mathrm{E}.$ of the system is$3$
3. The source which has provided the additional $\mathrm{K}.\mathrm{E}.$is the work done by the man to bring the brick closer to the body.

Step-by-step solution

Given

1. The angular speed of the platform,$\text{ω}=1.2\text{rev}/\text{s}$
2. The initial rotational inertia,${\text{I}}_{\text{i}}=6.0\text{kg}.{\text{m}}^2$
3. The new rotational inertia of the system is,${\text{I}}_{\text{f}}=2.0\text{kg}.{\text{m}}^2$

To understand the concept

Using the law of conservation of angular momentum, find the new angular speed of the system. Then using this, find the rotational new K.E of the system. Finally, using the rotational K.E. of the system, we can find the ratio of initial and new K.E.

Formula:

$\mathbf{K}\mathbf{.}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

The law of conservation of angular momentum,${\mathbf{I}}_{\mathbf{i}}{\mathbf{\omega }}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{I}}_{\mathbf{f}}{\mathbf{\omega }}_{\mathbf{f}}$ .

 Calculate the resulting angular speed of the platform

(a)

According to the law of conservation of angular momentum,

${\text{I}}_{\text{i}}{\text{ω}}_{\text{i}}={\text{I}}_{\text{f}}{\text{ω}}_{\text{f}}$

So, the new angular speed of the system is,

$\begin{array}{c}{\text{ω}}_{\text{f}}=\frac{{\text{I}}_{\text{i}}{\text{ω}}_{\text{i}}}{{\text{I}}_{\text{f}}}\\ \Rightarrow {\text{ω}}_{\text{f}}=\frac{6\times 1.2}{2}\\ \Rightarrow {\text{ω}}_{\text{f}}=3.6\mathrm{rev}/\mathrm{s}\end{array}$

Therefore, the angular speed of the system after brick is moved closer to the body of the man is ${\text{ω}}_{\text{f}}=3.6\mathrm{rev}/\mathrm{s}$.

Calculate the ratio of the new kinetic energy of the system to the original kinetic energy

(b)

The ratio of the new kinetic energy to the initial $K.E.$of the system is,

$\begin{array}{c}\frac{\text{K}.{\text{E}}_{\text{f}}}{\text{K}.{\text{E}}_{\text{i}}}=\frac{\frac{1}{2}{\text{Iω}}_{\text{f}}^2}{\frac{1}{2}{\text{Iωi}}_{\text{i}}^2}\\ \Rightarrow \frac{\text{K}.{\text{E}}_{\text{f}}}{\text{K}.{\text{E}}_{\text{i}}}=\frac{\frac{1}{2}\left(2\right){\left(3.6\right)}^2}{\frac{1}{2}\left(6\right){\left(1.2\right)}^2}\\ \Rightarrow \frac{\text{K}.{\text{E}}_{\text{f}}}{\text{K}.{\text{E}}_{\text{i}}}=3\end{array}$

Find the source which provided the added kinetic energy

(c)

After pulling the brick closer to the body, the man decreases the rotational inertia of the system which results in the increase in the angular speed. This work done by the man for pulling the brick closer has provided the additional$\mathrm{K}.\mathrm{E}.$