How to find speed of rotation of gears from kinetic energy of falling body?

Question

A pulley is fitted to the first driven shaft of compund gear train. Last driven shaft of gear train(gear 8) is fitted to the 12V DC generator (6W).Now to the rope extension at the one end of the pulley a mass of 10kg is added. The whole assembly is placed at an height of 2m from ground. The pulley is of 0.1m diameter.

The compound gear train consists of 8 spur gears of 2mm module having teeths Gear1 = 120, Gear2 = 110, Gear3 = 100, Gear4 = 40, Gear5 = 24, Gear6 = 22, Gear7 = 18, Gear8 = 18.

Gear 1&5, 2&6, 3&7, 4&8 are engaging. The gears 2&5, 3&6, 4&7 are on same shaft(concentric gears). Pulley and Gear 1 are on same shaft. DC generator and gear 8 are on same shaft. See fig. below.

All gears have Module = 2mm & involute tooth profile with pressure angle 20 deg. Space width of gears = 18mm.

I need to find the speed (RPM) at which Gear 1 i.e of 120 teeth and the pulley rotates as the mass of 10kg comes down due to gravity.

I know to solve the problem is a lot to ask but some info on how to solve would be helpful.

Again I am not asking to solve the problem and give me the answer. I have already solved it. I am just looking for different ways of solving it.

This is what i exactly need to know: I found that the mass when added to the pulley ropes with a load on generator, falls down slower with a constant velocity as compared to free fall condition where it falls with acceleration due to gravity. How is that possible and what are all the factors contributing to this change in type of motion?

This isnt my homework. Its for a hobby project of mine.

Thank you.

Answer 1

You have five wheels, and their rotational velocities can be written down as a function of the velocity of the big weight dropping. If we write that velocity as $v$, then the angular velocity of wheel 1 is $\omega_1 = \frac{v}{r_{pulley}}$, and the ratios of velocities of the other wheels follows the gear ratios, so

$$\omega_2 = \frac{N_1}{N_5}\omega_1\\ \omega_3 = \frac{N_2}{N_6}\omega_2$$

etcetera.

Finally, you can write down the total energy of the system (before any energy is extracted by the 12 V generator) as

$$E_{kinetic}= \frac12(mv^2 + I_1\omega_1^2 + I_2\omega_2^2+...)$$

When you express every $\omega$ in terms of $v$, you will discover that there is an "effective inertia" of the system that is much greater than $m$. Meaning that when the weight drops under gravity, its acceleration will be less than $g$ - while $mgh = \frac12 m_{eff}v^2$ where $m_{eff}$ is the effective mass which you will arrive at by following the above analysis through its conclusion.

Answer 2

I will give you an overall principle which you can use to solve your problem: The tangential speeds of gears in tooth-to-tooth contact must be the same. The angular speeds of gears mounted on a common axis must be the same.

$$v_{tan}=\omega r$$

That's all you need. You can do the arithmetic.