Answer to Question #124816 in Classical Mechanics for Cole Shotkoski

Question

A mechanical process transforms rotational motion and potential energy into thermal energy.

In the start of this process, an object of mass 2kg and momentum of inertia 2km/m2 moves at an initial angular speed of of 12Hz, undergoes a displacement of 5m [Down] while being subject to Earth’s normal gravity 9.8m/s2 [down]. Through this process, 20 J of thermal energy is produced.

a. Determine the FINAL ANGULAR SPEED of this process assuming the energy is conserved in this process (100% efficient)

b. How efficient would this process be at converting just potential energy into thermal energy, assuming that the potential energy and thermal energy remain unchanged?

Answer 1

As per the given question,

Let initially, object was at height $h_1$ and finally it was at height $h_2$ and final angular speed $\omega_2$ ​

Mass of the object $(m)=2kg$

Moment of inertia of the object $(I)=2kg-m^2$

Frequency of rotation$(f)=12Hz$

Angular velocity of the rotation $(\omega)=2\pi f=24\pi$

Vertical displacement$(h_1-h_2)=5m$

Gravitational acceleration $(g)=9.8m/sec^2$

Thermal energy $(H)=20J$

Now, applying the conservation of energy,

$\Rightarrow mgh_1+\frac{I\omega^2}{2}=mgh_1+\frac{I\omega_2^2}{2}+H$

$\Rightarrow \frac{I\omega_2^2}{2}=mg(h_1-h_2)+\frac{I\omega^2}{2}-H$

$\Rightarrow \frac{I\omega_2^2}{2}=5\times 9.8\times5J+\frac{2(24\pi)^2}{2}J-20J$

$\Rightarrow \frac{2\omega_2^2}{2}=5909.9 (r/s)^2$

$\Rightarrow \omega_2=\sqrt{5909.9}r/s=76.87 r/s$

b) When object was at certain height at that height it's rotational speed was $24\pi$ rad/sec but as it come down 5 below, it's rotational speed is increasing till $24.5\pi$ and heat is getting produce. means at the end, total stored potential energy is not getting convert into the heat, some of part of the change in potential energy is getting convert into the heat and remaining part of the energy is used in the increasing the of rotational energy, so it's efficiency is not very well.