If there are 2 gears meshed together and they are of different sizes, then rotating the smaller one will make the larger one spin with a smaller angular velocity but with more torque. And the opposite happens when you spin the larger one. Using a lower gear ratio in a bicycle for example, makes it easier to go uphill. How does the increased torque from the lower gear ratio help in this? Like how does the higher torque equate to a greater force to move the bike forward?
[Physics] Relation between force and torque for a set of gears/bicycle
forceshomework-and-exercisesnewtonian-mechanicsrotational-dynamicstorque
Related Solutions
The power input is roughly constant (that of a car is dictated by the total engine power while for a bicycle it depends on the user). The gear or similar tools adjusts the mechanical advantage so that a low gear will express the engine power in force rather than speed (recall that power is force times speed). On higher gears the force is traded in for speed.
Consider the diagram below:
Shown are crank, chain wheel, chain, sprocket and rear wheel (drive wheel).
Assume a constant force $F$ is applied to the end of the crank the question then is how does this result in an acceleration $a$ and what's the magnitude of $a$?
The force $F$ causes a torque $T$ about the centre of the crank:
$$T=FR_c$$
Now we can calculate the tension in the chain $F_c$:
$$F_c=\frac{T}{R_L},$$
with $R_L$ the radius of the chain wheel.
So:
$$F_c=\frac{R_c}{R_L}F$$
This causes torque about the sprocket:
$$T_S=F_cR_S$$
$$T_S=\frac{R_cR_S}{R_L}F$$
Assuming enough friction between the rear wheel contact and the floor to prevent slippage we can now calculate the drive force $F_D$:
$$F_D=\frac{T_S}{R_W}$$
So:
$$F_D=\frac{R_cR_S}{R_WR_L}F$$
Where:
$$\frac{R_cR_S}{R_WR_L}$$
is known as the gain ratio, the factor that when multiplied with the input force gives the drive force.
Assuming $m$ to be the mass of bicycle and rider combined then with Newton:
$$a=\frac{F_D}{m}.$$
Best Answer
Changing to an easier gear reduces the ratio of torque required at the crankset over torque at the wheel. Given constant torque at the wheel (riding up a hill of a given grade at a given speed) this means the torque, and pedal forces, required at the crankset are lower. The rider thus feels it is easier to turn the pedals.