Chapter 4, Problem 2PS

To determine

Find the angle through which the free wheel turns for each gear.

Answer to Problem 2PS

  $\boxed{480°,780°,990°,1152°,1425°}$

Explanation of Solution

Given information:

A bicycle’s gear ratio is the number of times the free wheel turns for every one turn of the chain wheel (see figure). The table shows the numbers of teeth in the free wheel and chain wheel for the first five gears of an $18$ -speed touring bicycle. The chain wheel completes one rotation for each gear. Find the angle through which the free wheel turns for each gear. Give your answers in both degrees and radians.

  $\begin{array}{lll}GearNumber\hfill & Numberofteethinfreewheel\hfill & Numberofteethinchainwheel\hfill \\ 1\hfill & 32\hfill & 24\hfill \\ 2\hfill & 26\hfill & 24\hfill \\ 3\hfill & 22\hfill & 24\hfill \\ 4\hfill & 32\hfill & 40\hfill \\ 5\hfill & 19\hfill & 24\hfill \end{array}$

Calculation:

Here, we will consider the gear ratio bicycle. This is the number of times the free wheel turns for every one turns of the chain wheel.

We will consider the chain wheel and the free wheel attached as below:

  

Now, the gear numbers and the number of teeth of both the wheels are mentioned in the table below:

  $\begin{array}{lll}GearNumber\hfill & Numberofteethinfreewheel\hfill & Numberofteethinchainwheel\hfill \\ 1\hfill & 32\hfill & 24\hfill \\ 2\hfill & 26\hfill & 24\hfill \\ 3\hfill & 22\hfill & 24\hfill \\ 4\hfill & 32\hfill & 40\hfill \\ 5\hfill & 19\hfill & 24\hfill \end{array}$

Now, we will consider the gear number $1$ . Here, for one rotation of chain wheel the free wheel will finish one rotation.

Thus, consider the radius of chain wheel $r_1$ and free wheel $r_2$ . They will travel the same distance as the chains are connected. So,

  $\begin{array}{l}2\pi r_1=\left(2\pi r_2\right)\\ r_1=r_2\end{array}$

Now, the teeth in chain wheel are $24$ . Thus, for each movement of teeth, the distance travelled is:

  $\frac{2\pi r_1}{24}$

Now, assume that the angle of rotation for free wheel is $\theta$ . So, for each movement of teeth the distance travelled is:

  $\frac{\theta r_2}{32}$

Now, these two are same:

  $\frac{2\pi r_1}{24}=\frac{\theta r_2}{32}$

So, put $r_1=r_2$ :

  $\begin{array}{l}\frac{2\pi \left(r_2\right)}{24}=\frac{\theta r_2}{32}\\ \theta =\frac{32\times 2\pi }{24}\\ \theta =\frac{32\pi }{12}\\ \theta =\frac{8\pi }{3}\end{array}$

Hence, the angle of rotation in radian is $\boxed{\frac{8\pi }{3}}$ .

Now, the relation between degree and radian is:

  $1°=\frac{\pi }{180}$

We will convert the angle into radian:

  $\begin{array}{l}\theta =\frac{8\pi }{3}\times \frac{1}{\frac{\pi }{180°}}\\ \theta =\frac{8\times 180°}{3}\\ \theta =480°\end{array}$

Hence, the angle in degrees is $\boxed{480°}$ .

Now, we will consider the gear number $2$ . Here, for one rotation of chain wheel the free wheel will finish two rotations.

Thus, consider the radius of chain wheel $r_1$ and free wheel $r_2$ . They will travel the same distance as the chains are connected. So,

  $\begin{array}{l}2\pi r_1=2\left(2\pi r_2\right)\\ r_1=2r_2\end{array}$

Now, the teeth in chain wheel are $24$ . Thus, for each movement of teeth, the distance travelled is:

  $\frac{2\pi r_1}{24}$

Now, assume that the angle of rotation for free wheel is $\theta$ . So, for each movement of teeth the distance travelled is:

  $\frac{\theta r_2}{26}$

Now, these two are same:

  $\frac{2\pi r_1}{24}=\frac{\theta r_2}{26}$

So, put $r_1=2r_2$ :

  $\begin{array}{l}\frac{2\pi \left(r_2\right)}{24}=\frac{\theta r_2}{26}\\ \theta =\frac{26\times 4\pi }{24}\\ \theta =\frac{26\pi }{6}\\ \theta =\frac{13\pi }{3}\end{array}$

Hence, the angle of rotation in radian is $\boxed{\frac{13\pi }{3}}$ .

Now, the relation between degree and radian is:

  $1°=\frac{\pi }{180}$

We will convert the angle into radian:

  $\begin{array}{l}\theta =\frac{13\pi }{3}\times \frac{1}{\frac{\pi }{180°}}\\ \theta =\frac{13\times 180°}{3}\\ \theta =780°\end{array}$

Hence, the angle in degrees is $\boxed{780°}$ .

Now, we will consider the gear number $3$ . Here, for one rotation of chain wheel the free wheel will finish two rotations.

Thus, consider the radius of chain wheel $r_1$ and free wheel $r_2$ . They will travel the same distance as the chains are connected. So,

  $\begin{array}{l}2\pi r_1=3\left(2\pi r_2\right)\\ r_1=3r_2\end{array}$

Now, the teeth in chain wheel are $24$ . Thus, for each movement of teeth, the distance travelled is:

  $\frac{2\pi r_1}{24}$

Now, assume that the angle of rotation for free wheel is $\theta$ . So, for each movement of teeth the distance travelled is:

  $\frac{\theta r_2}{22}$

Now, these two are same:

  $\frac{2\pi r_1}{24}=\frac{\theta r_2}{22}$

So, put $r_1=3r_2$ :

  $\begin{array}{l}\frac{2\pi \left(3r_2\right)}{24}=\frac{\theta r_2}{22}\\ \theta =\frac{26\times 6\pi }{24}\\ \theta =\frac{22\pi }{4}\\ \theta =\frac{11\pi }{2}\end{array}$

Hence, the angle of rotation in radian is $\boxed{\frac{11\pi }{2}}$ .

Now, the relation between degree and radian is:

  $1°=\frac{\pi }{180}$

We will convert the angle into radian:

  $\begin{array}{l}\theta =\frac{11\pi }{3}\times \frac{1}{\frac{\pi }{180°}}\\ \theta =\frac{11\times 180°}{2}\\ \theta =990°\end{array}$

Hence, the angle in degrees is $\boxed{990°}$ .

Now, we will consider the gear number $4$ . Here, for one rotation of chain wheel the free wheel will finish two rotations.

Thus, consider the radius of chain wheel $r_1$ and free wheel $r_2$ . They will travel the same distance as the chains are connected. So,

  $\begin{array}{l}2\pi r_1=4\left(2\pi r_2\right)\\ r_1=4r_2\end{array}$

Now, the teeth in chain wheel are $40$ . Thus, for each movement of teeth, the distance travelled is:

  $\frac{2\pi r_1}{40}$

Now, assume that the angle of rotation for free wheel is $\theta$ . So, for each movement of teeth the distance travelled is:

  $\frac{\theta r_2}{32}$

Now, these two are same:

  $\frac{2\pi r_1}{40}=\frac{\theta r_2}{32}$

So, put $r_1=4r_2$ :

  $\begin{array}{l}\frac{2\pi \left(4r_2\right)}{40}=\frac{\theta r_2}{32}\\ \theta =\frac{32\times 8\pi }{40}\\ \theta =\frac{32\pi }{5}\end{array}$

Hence, the angle of rotation in radian is $\boxed{\frac{32\pi }{5}}$ .

Now, the relation between degree and radian is:

  $1°=\frac{\pi }{180}$

We will convert the angle into radian:

  $\begin{array}{l}\theta =\frac{32\pi }{5}\times \frac{1}{\frac{\pi }{180°}}\\ \theta =\frac{32\times 180°}{5}\\ \theta =1152°\end{array}$

Hence, the angle in degrees is $\boxed{1152°}$ .

Now, we will consider the gear number $5$ . Here, for one rotation of chain wheel the free wheel will finish two rotations.

Thus, consider the radius of chain wheel $r_1$ and free wheel $r_2$ . They will travel the same distance as the chains are connected. So,

  $\begin{array}{l}2\pi r_1=5\left(2\pi r_2\right)\\ r_1=5r_2\end{array}$

Now, the teeth in chain wheel are $24$ . Thus, for each movement of teeth, the distance travelled is:

  $\frac{2\pi r_1}{24}$

Now, assume that the angle of rotation for free wheel is $\theta$ . So, for each movement of teeth the distance travelled is:

  $\frac{\theta r_2}{19}$

Now, these two are same:

  $\frac{2\pi r_1}{24}=\frac{\theta r_2}{19}$

So, put $r_1=5r_2$ :

  $\begin{array}{l}\frac{2\pi \left(5r_2\right)}{24}=\frac{\theta r_2}{19}\\ \theta =\frac{19\times 10\pi }{24}\\ \theta =\frac{95\pi }{12}\end{array}$

Hence, the angle of rotation in radian is $\boxed{\frac{95\pi }{12}}$ .

Now, the relation between degree and radian is:

  $1°=\frac{\pi }{180}$

We will convert the angle into radian:

  $\begin{array}{l}\theta =\frac{95\pi }{12}\times \frac{1}{\frac{\pi }{180°}}\\ \theta =\frac{95\times 180°}{12}\\ \theta =1425°\end{array}$

Hence, the angle in degrees is $\boxed{1425°}$ .