Bicycle gears: (a) How is the angular velocity\({{\bf{\omega }}_{\bf{R}}}\) of the rear wheel of a bicycle related to the angular velocity\({{\bf{\omega }}_{\bf{F}}}\)of the front sprocket and pedals? Let \({{\bf{N}}_{\bf{F}}}\) and \({{\bf{N}}_{\bf{R}}}\) be the number of teeth on the front and rear sprockets, respectively, Fig. 8–58. The teeth are spaced the same on both sprockets and the rear sprocket is firmly attached to the rear wheel. (b) Evaluate the ratio when the front and rear sprockets have 52 and 13 teeth, respectively, and (c) when they have 42 and 28 teeth.

When an object rotates about a fixed axis, thelinear velocity (v) of any point on the object located at a distance r from the axis of rotation is related to the angular velocity \(\left( \omega \right)\) of the object by the following relation:

\(v = r\omega \)

In this problem,the linear velocity of the chain is related to the angular velocity of the sprocket by the above expression.

The angular velocity of the rear sprocket is =\({\omega _{\rm{R}}}\).

The angular velocity of the front sprocket and pedals is =\({\omega _{\rm{F}}}\).

The number of teeth on the front sprocket is =\({N_{\rm{F}}}\).

The number of teeth on the rear sprocket is =\({N_{\rm{R}}}\).

Let \({R_{\rm{F}}}\) and \({R_{\rm{R}}}\) be the radii of the front and rear sprockets, respectively.

The spacing between the teeth in both the sprockets is equal. If this spacing is equal to d, then the product of this spacing and the number of teeth in a sprocket is equal to the circumference of the sprocket.

For the front sprocket,

\(\begin{aligned}{c}2\pi {R_{\rm{F}}} = d{N_{\rm{F}}}\\{R_{\rm{F}}} = \frac{{d{N_{\rm{F}}}}}{{2\pi }}\end{aligned}\) ... (i)

For the rear sprocket,

\(\begin{aligned}{c}2\pi {R_{\rm{R}}} = d{N_{\rm{F}}}\\{R_{\rm{R}}} = \frac{{d{N_{\rm{R}}}}}{{2\pi }}\end{aligned}\) ... (ii)

The linear speed of the chain when passing through the front sprocket is:

\({v_{\rm{F}}} = {R_{\rm{F}}}{\omega _{\rm{F}}}\)

The linear speed of the chain when passing through the rear sprocket is:

\({v_{\rm{R}}} = {R_{\rm{R}}}{\omega _{\rm{R}}}\)

Since the same chain runs over both front and rear sprockets, its linear velocity must remain the same, i.e.,

\(\begin{aligned}{c}{v_{\rm{F}}} = {v_{\rm{R}}}\\{R_{\rm{F}}}{\omega _{\rm{F}}} = {R_{\rm{R}}}{\omega _{\rm{R}}}\\\frac{{{\omega _{\rm{R}}}}}{{{\omega _{\rm{F}}}}} = \frac{{{R_{\rm{F}}}}}{{{R_{\rm{R}}}}}\end{aligned}\)

On substituting the values of \({R_{\rm{F}}}\)and \({R_{\rm{R}}}\)from equations (i) and (ii) in the above expression, you will get:

\(\begin{aligned}{c}\frac{{{\omega _{\rm{R}}}}}{{{\omega _{\rm{F}}}}} = \frac{{d{N_{\rm{F}}}}}{{2\pi }} \times \frac{{2\pi }}{{d{N_{\rm{R}}}}}\\\frac{{{\omega _{\rm{R}}}}}{{{\omega _{\rm{F}}}}} = \frac{{{N_{\rm{F}}}}}{{{N_{\rm{R}}}}}\end{aligned}\)

Thus, the angular velocity of the rear wheel of the bicycle is related to the angular velocity of the front sprocket by the above expression.

The ratio of the angular velocities of the rear sprocket to the front sprocket when \({N_{\rm{F}}} = 52\) and \({N_R} = 13\)is:

\(\begin{aligned}{c}\frac{{{\omega _{\rm{R}}}}}{{{\omega _{\rm{F}}}}} = \frac{{{N_{\rm{F}}}}}{{{N_{\rm{R}}}}}\\ = \frac{{52}}{{13}}\\ = 4\end{aligned}\)

Thus, the ratio of the angular velocity of the rear sprocket to the angular velocity of the front sprocket is \(4:1\).

The ratio of the angular velocities of the rear sprocket to the front sprocket when \({N_{\rm{F}}} = 52\) and \({N_R} = 13\)is:

\(\begin{aligned}{c}\frac{{{\omega _{\rm{R}}}}}{{{\omega _{\rm{F}}}}} = \frac{{{N_{\rm{F}}}}}{{{N_{\rm{R}}}}}\\ = \frac{{42}}{{28}}\\ = \frac{3}{2}\end{aligned}\)

Thus, the ratio of the angular velocity of the rear sprocket to the angular velocity of the front sprocket is 3:2.