Based on the concept of the moment of inertia, my book gives the following relationship between the size of gear and amount of torque supplied to the drive wheel:
Because it [outermost gear] had has the largest radius, this gear provides the largest torque to the drive wheel. A large torque is required initially, because the bicycle starts from rest. As the bicycle rolls faster, the tangential speed of the chain increases, eventually becoming too fast for the cyclist to maintain by pushing the pedals. The chain is then moved to a gear with a smaller radius, so the chain has a smaller tangential speed that the cyclist can more easily maintain.
The first sentence I understand. According to
\begin{equation}
\tau = I\alpha = (mr^2)\alpha,
\end{equation}
the bigger the radius of the gear means a bigger moment of inertia, I, and so requires more torque to give the drive/rear wheel the same angular acceleration. However, I am not clear what "m" here corresponds to on the gear. Is the mass of the point on that specific gear in contact with the chain?
For the 2nd sentence, if the change of velocity, how does initial velocity = 0 matter as to the torque needed?
The rest I am pretty sure. But feel free to check my interpretation and point out any mistake:
At a certain point in time, the cyclist stop exerting torque on the drive/rear wheel. Assume no friction and a constant angular velocity of the drive wheel. Because the sieve wheel is soldered to all gears, they have the same angular velocity at all times. The bigger the radius of the gear and bigger the tangential velocity of the gear/wheel/chain.
One thing though, how does tangential velocity of the gear/chain has anything to do with how hard it is for the cyclist to maintain the tangential velocity? I thought for bikes these days, you can just out your feet on the pedal and don't have to let it move with the rotation of the wheel。
Best Answer
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:
$$a=\frac{F_D}{m}.$$
(Source.)