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I am thinking of two circular gears with radii $R_1$ and $R_2=2R_1$. Gear 1 initially spins at $\Omega_1=10$ $rpm$ while the Gear 2 is at rest. After the gears come into contact, Gear 2 angularly accelerates and eventually reaches the steady speed $\Omega_2=20$ $rpm$.

Gear 1 applies a torque $\tau_1=F_1 R_1$ on the second gear, correct? Why doesn't torque $\tau_1$ continue to increase the speed of the Gear 2? Why does $\Omega_2$ stop at 20 rpm? After all, an unbalanced torque should continue to provide angular acceleration...

Angular speed is amplified from $10$ $rpm$ to $20$ $rpm$ but the torques are $\tau_2<\tau_1$. Is torque $\tau_1=R_1F_1$ the torque acting on Gear 1 or on Gear 2? Is the force $F_1$ acting on Gear 1 to keep it spinning at $\Omega_1$? Or is $F_1$ the force acting on Gear 2 to make/keep it spinning at $\Omega_2$? I am confused.

Is $\tau_2=R_2F_2$ the torque that Gear 2 could apply to a 3rd hypothetical gear, if it came in contact with it, or is it the torque acting on Gear 2 itself?

Thank you

John Alexiou
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Brett Cooper
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  • The gears exchange momentum through their contact, and the change in speed is instantaneous. There is no measure of torque in this sense. – John Alexiou Oct 24 '20 at 02:28
  • The question is similar and the answer applies to both questions. – John Alexiou Oct 24 '20 at 02:30
  • Thank you John. I will go over those questions/answers. What threw me off is that gear trains are used to amplify/minify either torque or angular speed. Our discussion involves a change in angular speed. The purpose of torque magnification would be to use that larger torque for a specific purpose, like raising a load more easily... – Brett Cooper Oct 24 '20 at 20:23

2 Answers2

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Why doesn't torque $\tau_1$ continue to increase the speed of the Gear 2?

Forces and torques appear when the two disks have relative motion. As gear 2 accelerates (and gear 1 decelerates), eventually they reach a point where their speed at the point of contact is matched, there is no relative motion.

It's similar to dropping a package on a conveyor belt. There is a force that appears (via friction) that accelerates the package. But once the package matches the speed of the belt, the force disappears and the acceleration stops. The same thing happens between the disks.

The forces between the gears are identical. If you imagine gear1 is getting a 10N push from gear2, then gear2 is getting a 10N push from gear1. $|F_1| = |F_2|$

The torque felt by the disks due to this equal force will be different due to the radius. The larger disk will feel the larger torque.

But the torque $\tau_2$ does not disappear but stops accelerating gear 2...Why?

I don't understand the question. When the gears stop sliding (they roll with same tangential speed), the friction disappears. There is no force, there is no torque. When the torque is gone, there is no acceleration. Why do you think the torque doesn't disappear?

A third gear placed in contact with gear 2 will share the same force $F$ and have a torque that depends on its radius $R_3$

It will? That seems unwarranted.

Any pair of disks will share a force couple. If disk1 touches disk2, there will be equal magnitude forces due to this touch on both disks.

If disk2 touches disk3, there will be equal magnitude forces on both disks. But they may be different from the force pair between 1 and 2.

BowlOfRed
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  • I see, so the forces are the same, $R_2>R_1$ hence torque $\tau_2>\tau_1$. Torque $\tau_2$ is acting on and speeding up gear 2 until the final velocity of $5rpm$. But the torque $\tau_2$ does not disappear but stops accelerating gear 2...Why?

    A third gear placed in contact with gear 2 will share the same force $F$ and have a torque that depends on its radius $R_3$. I think I get it :)

     – Brett Cooper Oct 23 '20 at 20:58
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I think there is something wrong with the angular velocities you have specified. After gears come into contact, assuming they are not "skipping" or "sliding", the condition $\Omega_1R_1=\Omega_2R_2$ must hold. This doesn't hold with $\Omega_1=10\ \text{rpm}$ and $\Omega_2=20\ \text{rpm}$. I will assume in the rest of this answer that the final angular velocity of gear 2 is $\Omega_2=5\ \text{rpm}$.

In order for gear 2 to reach $\Omega_2 = 5\ \text{rpm}$, there needs to be an additional torque driving gear 1 and keeping it rotating at $\Omega_1 = 10\ \text{rpm}$. Otherwise (even in the absence of friction continuously slowing down the gears), gear 1 coming into contact with gear 2 will slow gear 1 down, as angular momentum would need to be conserved, so you would have the final gear angular velocities less than $10\ \text{rpm}$ and $5\ \text{rpm}$, respectively.

Assuming $F_1$ is the magnitude of the (tangential) force applied by gear 1 on gear 2 and $F_2$ is the magnitude of the (tangential) force applied by gear 2 on gear 1, then by Newton's third law, these forces are equal in magnitude, i.e. $F_1=F_2=F$. Then, $\tau_1=FR_1$ is the torque acting to slow down gear 1 and $\tau_2=FR_2$ is the torque acting to speed up gear 2.

As gear 1 is accelerating gear 2, $\tau_2$ is non-zero. Meanwhile $\tau_1$ is also non-zero, so in the absence of an external torque on gear 1, it would decelerate. After the gears reach their final velocities, there is no net kinetic friction or (tangential) contact force acting on either gear, i.e. $F=0$, so there is no net torque on either gear.

Puk
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  • Thanks. In general mechanical situations, Gear 1 is driven by a spinning central shaft which keeps it spinning at a constant angular velocity $\Omega_1$. Gear 2 (with sprockets) is generally already in contact with Gear 1 and spins at its lower $\Omega_2$. As you guys mention, the relative speed is always zero at the contact point if the gears don't slip. The torques nonzero are different, while the forces equal and opposite, until the torques disappearing, I am referring to your statement "After the gears reach their final velocities,......, F=0, so there is no net torque on either gear." – Brett Cooper Oct 24 '20 at 13:39