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Physics / Physics: Principles with Applications 6 / Chapter 8 / Problem 73GP

Physics: Principles with Applications | 6th Edition | ISBN: 9780321569837 | Authors: Douglas C. Giancoli

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Bicycle gears: (a) How is the angular velocity wR of the rear wheel of a bicycle related

Chapter 8, Problem 73GP

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QUESTION:

(a) For a bicycle, how is the angular speed of the rear wheel $\left({{\omega{}}_{R}}\right)$ related to that of the pedals and front sprocket $\left({{\omega{}}_{F}}\right)$ , Fig. $8-52?$ That is, derive a formula for ${\omega{}}_{R}/{\omega{}}_{F}$ . Let ${N}_{F}$ and ${N}_{R}$ be the number of teeth on the front and rear sprockets, respectively. The teeth are spaced equally on all sprockets so that the chain meshes properly. (b) Evaluate the ratio ${\omega{}}_{R}/{\omega{}}_{F}$ when the front and rear sprockets have 52 and 13 teeth, respectively, and $(c)$ when they have 42 and 28 teeth.

Figure 8-52 Problem 73

Questions & Answers

QUESTION:

(a) For a bicycle, how is the angular speed of the rear wheel $\left({{\omega{}}_{R}}\right)$ related to that of the pedals and front sprocket $\left({{\omega{}}_{F}}\right)$ , Fig. $8-52?$ That is, derive a formula for ${\omega{}}_{R}/{\omega{}}_{F}$ . Let ${N}_{F}$ and ${N}_{R}$ be the number of teeth on the front and rear sprockets, respectively. The teeth are spaced equally on all sprockets so that the chain meshes properly. (b) Evaluate the ratio ${\omega{}}_{R}/{\omega{}}_{F}$ when the front and rear sprockets have 52 and 13 teeth, respectively, and $(c)$ when they have 42 and 28 teeth.

Figure 8-52 Problem 73

ANSWER:

Solution 73GP

Step 1 of 3

Part a

We are required to compare the angular velocity of the rear wheel to that of the front wheel.

The linear speed of the rear wheel,

$v=R{\ }_{R}\omega{}{\ }_{R}$

The linear speed of the front wheel,

$v=R{\ }_{F}\omega{}{\ }_{F}$

Since both are equal,

$R{\ }_{R}\omega{}{\ }_{R}=R{\ }_{F}\omega{}{\ }_{F}$

${\frac{R{\ }_{R}}{R{\ }_{F}}=\frac{\omega{}{\ }_{F}}{\omega{}{\ }_{R}}}_{\ }$ …..(1)

Let the number of the front wheel be $N{\ }_{F}$ and that of the rear wheel be $N{\ }_{R}$ .

Let the separation between the consecutive teeth in each is $d$ .

Therefore, the circumference,

$2\pi{}R{\ }_{R}=N{\ }_{R}d$

$2\pi{}R{\ }_{F}=N{\ }_{F}d$

So, $\frac{R{\ }_{R}}{R{\ }_{F}}=\frac{N{\ }_{R}}{N{\ }_{F}}$ …..(2)

From equations (1) and (2),

$\frac{\omega{}{\ }_{F}}{\omega{}{\ }_{R}}=\frac{N{\ }_{R}}{N{\ }_{F}}$

$\frac{\omega{}{\ }_{R}}{\omega{}{\ }_{F}}=\frac{N{\ }_{F}}{N{\ }_{R}}$

Therefore, the angular velocity of a wheel is inversely proportional to the number of teeth in that wheel.

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