Rotational Kinetic Energy Conservation

Question

Consider a chain around two gears, one of of radius $r_1$ and the other of $r_2$. Say the gear $r_1$ is attached to a rotational device that delivers torque $\tau$ . After a quarter cycle of rotation you have input energy $E = \tau\cdot\pi/4$ rotational energy into the system.

Say you have the same system but this time gear $r_2$ is replaced with another gear of radius $r_3$ where $r_3>r_2$. Again you spin it with torque $ \tau$ for a quarter cycle so you have the same energy in the system.

My questions are:

1. Would system 2 (with gear $r_3$) be spinning faster than system 1 (with gear $r_2$)? I think it would since there is a larger gear.

2. If it is spinning faster, how is that justifiable? You input the same energy into both the systems but one is spinning faster than the other.

Thanks for any help.

Edit for clarification:

1. I'm asking about the angular velocity of the first gear in both systems
2. The rotational device is concentrically connected to the first gear

Answer 1

Let's make some simplifying assumptions here:

1. The gears are much lighter than the chain, so we can assume all of the mass is located on the outside of the gears in the chain itself.

2. The chains wrap all the way around the gear. This is probably less realistic, but this way we can treat the system as two thin hoops that are constrained to spin at the same linear velocity. I don't think this messes up the overall analysis.

3. The chain has a uniform linear mass density $\lambda$.

Therefore, a gear of radius $R$ will have a mass of $m=2\pi R\lambda$ and a moment of inertia of $I=mR^2=2\pi R^3\lambda$ Additionally, given the constraint of the gears being connected by the chain, it must be that the gears have the same linear velocity $v=\omega_1R_1=\omega_2R_2$ at their edges.

The kinetic energy of the two-gear system will then be

$$K=\frac12I_1\omega_1^2+\frac12I_2\omega_2^2=\pi\lambda R_1^2(R_1+R_2)\omega_1^2$$

So as you can see, for the same amount of work, the larger $R_2$ is, the smaller $\omega_1$ will be. Therefore, the larger the second gear the slower everything will rotate.

Answer 2

Assuming no other stuff the larger system spins slower due to its larger moment of inertia.

So like the total kinetic energy in the first system assuming a massless chain is given by the angular velocities $\omega_{1,2}$ as $$ K=\frac12 I_1\omega_1^2 +\frac12 I_2\omega_2^2 $$where the moments of inertia are $I_{1,2}.$ The chain between them forces $r_1\omega_1=r_2\omega_2$ when it is taut, so this is $$K =\frac12\left(I_1(r_2/r_1)^2+I_2\right)\omega_2^2$$ and if the gears are of similar construction (same thickness material but pattern otherwise zoomed-in from one size to the other) you'll have $I_1 =(r_1/r_2)^4 I_2$ from dimensional analysis alone, so you will have a rate $$\omega_1=\frac{r_2}{r_1} ~\frac{\tau\cdot\pi/4}{I_2(1+(r_1/r_2)^2)}.$$

So if $s=r_{1,3}/r_2$ you have a term that goes like $1/(s +s^3),$ it decreases as $s$ increases. Bigger systems spin slower with the same energy.

Answer 3

We get the energy equation $$\tau \cdot \theta = \frac {I_1 \omega_1^2}{2} + \frac {I_2 \omega_2^2}{2}$$.

As the two gears are attatched to each other, their linear speed at the edges is same. So, we get the equation $\omega_1 r_1 = \omega_2 r_2$.

Let the moment of inertia of a gear be $I = k m r^2$. Assuming that the gears are made of the same material, their 2-D density - $\sigma$ (mass per unit area) is constant. So, $I = k (\sigma \pi r^2) r^2$, i.e. $I = k' r^4$

Substituting $r_1 = \frac {\omega_2 r_2}{\omega_1}$ and $I_n = k' r_n^4$ in the energy equation, we get $$ \tau \cdot \theta = \frac {k' \omega_2^2 r_2^2}{2} (r_1^2 + r_2^2)$$ As $\tau \cdot \theta $ is constant in both cases, $$ \omega_2 = \frac {k''}{r_2 \sqrt{r_1^2 + r_2^2}}$$ So, when $r_2$ is increased to $r_3$, it is apparent from the equation that $\omega_2$ will reduce to $\omega_3$.

The bigger gear will only spin faster if its density is sufficiently lower than the smaller gear to reduce its moment of inertia.