Question

The short and long hand of a clock are $4\mathrm{cm}$ and $6\mathrm{cm}$ respectively. Find the sum of distances travelled by their tips in one day.

Answer 1

Step 1: Given information

Length of short hand $r_1=4\mathrm{cm}$.

Length of long hand $r_2=6\mathrm{cm}$.

Step 2: Determine the distance travelled by the short hand (hour hand) in one day.

The circumference of a circle $\left(C\right)$ is calculated by $C=2\mathrm{\pi r}$ where $r$ represents the radius of the circle.

Circumference of the path travelled by the short hand $C_1=2{\mathrm{\pi r}}_1$.

$\begin{array}{rcl}& \Rightarrow & C_1=2\times \mathrm{\pi }\times 4\\ & =& 8\mathrm{\pi }\mathrm{cm}\end{array}$

The short hand of a clock completes $2$ rotation in $1$ day $\left(24\mathrm{hr}\right)$.

Distance travelled by short hand in one day

$\begin{array}{rcl}& =& 2\times C_1\\ & =& 2\times 8\mathrm{\pi }\\ & =& 16\mathrm{\pi }\mathrm{cm}\end{array}$

Step 3: Determine the distance travelled by the long hand (minute hand) in one day.

Circumference of the path travelled by the long hand $C_2=2{\mathrm{\pi r}}_2$.

$\begin{array}{rcl}& \Rightarrow & C_2=2\times \mathrm{\pi }\times 6\\ & =& 12\mathrm{\pi }\mathrm{cm}\end{array}$

The minute hand makes $1$ revolution in one hour.

In $1$ day, it makes $24$ revolutions.

Distance travelled by short hand in one day

$\begin{array}{rcl}& =& 24\times C_2\\ & =& 24\times 12\mathrm{\pi }\\ & =& 288\mathrm{\pi }\mathrm{cm}\end{array}$

Step 4: Determine the total distances travelled by short hand and long hand in one day.

$\begin{array}{rcl}\mathrm{Total}\mathrm{distance}& =& 16\mathrm{\pi }+288\mathrm{\pi }\\ & =& 304\mathrm{\pi }\mathrm{cm}\end{array}$

The value of $\mathrm{\pi }$ is $3.14$.

$\begin{array}{rcl}\mathrm{Total}\mathrm{distance}& =& 304\times 3.14\\ & =& 954.56\mathrm{cm}\end{array}$

Hence, the sum of distances travelled by short hand and long hand in one day is $304\mathrm{\pi }\mathrm{cm}or954.56\mathrm{cm}$.