Question

II The minute hand on a watch is$2.0\text{cm}$in length. What is the displacement vector of the tip of the minute hand

a. From \(8: 00\) to \(8: 20\) A.M.?

b. From \(8: 00\) to \(9: 00\) A.M.?

Short answer

a. The displacement vector of the tip of minute hand of watch is $\Delta \overrightarrow{l}=\sqrt{3}\widehat{i}-3\widehat{j}\text{cm}$.

b. The displacement vector of the tip of minute hand of watch is$\Delta \overrightarrow{l}=\overrightarrow{0}\text{cm}$.

Step-by-step solution

Part. (a).

If $/$ is the length of minute hand of a watch representing the magnitude of a vector making an angle $\theta$anticlockwise with the positive direction of $x$-axis.

Vector $\overrightarrow{l}$can be represented in terms of its $x$and $y$components as,

$\overrightarrow{l}=l_x\widehat{i}+l_y\widehat{j}$

Where its $x$and $y$components are,

$\overrightarrow{l}=l_x\widehat{i}+l_y\widehat{j}$

Where its $x$and $y$components are,

$\begin{array}{l}{l}_x=l\cos \theta \\ l_y=l\sin \theta \end{array}$

Substitute $l\cos \theta$for $l_x$and $l\sin \theta$for $l_y$in equation (1) position vector of tip of minute hand of watch can be written as,

$\overrightarrow{l}=l\cos \theta \widehat{i}+l\sin \theta \widehat{j}\dots \dots \left(2\right)$

If $\overrightarrow{l}$is initial position vector of tip of minute hand and ${\overrightarrow{l}}_f$is final position vector of tip of minute hand then displacement vector of $\Delta \overrightarrow{l}$tip of minute hand of watch

$\overrightarrow{\Delta l}=\overrightarrow{l_f}-{\overrightarrow{l}}_i\dots \dots \left(3\right)$

When minute hand is at its position is drawn as below.

Angle made by minute hand anticlockwise with the positive direction of $x$-axis is ${90}^{°}$.

Substitute $2\text{cm}$for / and ${90}^{°}$for in $\theta$equation (2).

Initial position vector of tip of minute hand

$\begin{array}{l}{\overrightarrow{l}}_i=(2\text{cm})\cos {90}^{°}\widehat{\text{i}}+(2\text{cm})\sin {90}^{°}\widehat{\text{j}}\\ {\overrightarrow{l}}_i=(2\text{cm})\widehat{\text{j}}\end{array}$

When minute hand is $8.20\text{AM}$at its position is drawn as below.

If watch is divided into 12 parts, each part makes an angle ${30}^{°}$,

Then angle made anticlockwise with the positive direction of $x$-axis by 11 parts is $\left(11\right)\left({30}^{°}\right)={330}^{°}$

So, angle made by minute hand anticlockwise with the positive direction of x-axis is ${330}^{°}$.

Substitute$2\text{cm}$for $/$ and${330}^{°}$for$\theta$

Final position vector of tip of minute hand

$\begin{array}{l}{\overrightarrow{l}}_f=(2\text{cm})\cos {330}^{°}\widehat{\text{i}}+(2\text{cm})\sin {330}^{°}\widehat{\text{j}}\\ {\overrightarrow{l}}_f=(\sqrt{3}\text{cm})\widehat{i}-(1\text{cm})\widehat{j}\end{array}$

Substituting values of $\overrightarrow{l_f}$and $\overrightarrow{l}$in equation (3).

$\begin{array}{l}\overrightarrow{\Delta l}={\overrightarrow{l}}_f-{\overrightarrow{l}}_i\\ \overrightarrow{\Delta l}=(\sqrt{3}\widehat{i}-\widehat{j})-\left(2\widehat{j}\right)\\ \Delta \overrightarrow{l}=(\sqrt{3}\text{cm})\widehat{\text{i}}-(3\text{cm})\widehat{\text{j}}\end{array}$

Therefore, the displacement vector of the tip of minute hand of watch is $\Delta \overrightarrow{l}=\sqrt{3}\widehat{i}-3\widehat{j}\text{cm}$.

Part.(b).

When minute hand is at $8.00\text{AM}$its position is drawn in blue color as below. (Red arrow represents hour hand)

Angle made by minute hand anticlockwise with the positive direction of x-axis is ${90}^{°}$.

Substitute $2\text{cm}$for / and ${90}^{°}$for $\theta$ initial position vector of tip of minute hand.

$\begin{array}{l}{\overrightarrow{l}}_i=(2\text{cm})\cos {90}^{°}\widehat{\text{i}}+(2\text{cm})\sin {90}^{°}\widehat{\text{j}}\\ {\overrightarrow{l}}_i=(2\text{cm})\widehat{\text{j}}\end{array}$

When minute hand is at $9.00\text{AM}$its position is drawn in blue color as below. . (Red arrow represents hour hand).

So, angle made by minute hand anticlockwise with the positive direction of x-axis is ${90}^{°}$.

Substitute $2\text{cm}$for / and ${90}^{°}$for $\theta$final position vector of tip of minute hand,

$\begin{array}{l}{\overrightarrow{l}}_f=(2\text{cm})\cos {90}^{°}\widehat{\text{i}}+(2\text{cm})\sin {90}^{°}\widehat{\text{j}}\\ {\overrightarrow{l}}_f=(2\text{cm})\widehat{\text{j}}\end{array}$

Substituting values of $\overrightarrow{l_f}$and $\overrightarrow{l}$in equation (3).

$\begin{array}{l}\overrightarrow{\Delta l}={\overrightarrow{l}}_f-{\overrightarrow{l}}_i\\ \Delta \overrightarrow{l}=(2\text{cm})\widehat{\text{j}}-(2\text{cm})\widehat{\text{j}}\end{array}$

Initial position of minute hand of a clock is in $+Y$direction and final position is also in $+Y$directic so displacement vector is a null vector because initial and final position vector are same.

Therefore, the displacement vector of the tip of minute hand of watch is$\Delta \overrightarrow{l}=\overrightarrow{0}\text{cm}$.