The minute hand of a wall clock measures${}^{\mathbf{10}\mathbf{}\mathbf{cm}}$ from its tip to the axis about which it rotates. The magnitude and angle of the displacement vector of the tip are to be determined for three time intervals. What are the (a) magnitude and (b) angle from a quarter after the hour to half past, the (c) magnitude and (d) angle for the next half hour, and the (e) magnitude and (f) angle for the hour after that?

The length of minute hand is 10 cm.

In a coordinate system, the position vector gives the location of an object relative to the origin. The displacement vector gives the change in the position of an object.

The expression for the displacement vector is given as:

$∆\overrightarrow{r}=\overrightarrow{r}-\overrightarrow{r_0}$ … (i)

Here, ${}^{\overrightarrow{r_0}}$and ${}^{\overrightarrow{r}}$are the position vectors for initial and final positions respectively.

The expression for the magnitude of the displacement is givenas:

$∆\overrightarrow{r}=\sqrt{{r_x}^2+{r_y}^2}$ … (ii)

Where $r_x$and $r_y$are the components of the vector.

The direction of vector is given as:

$\theta ={\tan }^{-1}\left(\frac{r_x}{r_y}\right)$ … (iii)

Consider the origin can be chosen as the center of the clock, the direction towards the right is the positive x direction and the direction upwards is the positive y direction.

Initially, the minute hand was along the positive x axis, i.e. in the direction of 3:00.

So, the position vectors are,

$\overrightarrow{r_0}=\left(10\mathrm{cm}\right)\hat{\mathrm{i}}\mathrm{and}\overrightarrow{\mathrm{r}}=\left(-10\mathrm{cm}\right)\hat{\mathrm{j}}$

Using equation (i), the displacement vector is,

$∆\overrightarrow{r}=\left(-10\mathrm{cm}\right)\hat{\mathrm{j}}=\left(-10\mathrm{cm}\right)\hat{\mathrm{i}}$

Using equation (ii), the magnitude of displacement is,

$$\begin{gathered} ∆\overrightarrow{r}=\sqrt{{\left(-10\mathrm{cm}\right)}^2+{\left(-10\mathrm{cm}\right)}^2} \\ =14.14\mathrm{cm} \end{gathered}$$

Thus, the magnitude of displacement is $14.14\mathrm{cm}.$

Using equation (iii), the direction of displacement will be,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-10\mathrm{cm}}{-10\mathrm{cm}}\right) \\ =45°or-135° \end{gathered}$$

Thus, the angle is$45°$from the negative X axis in counter clockwise direction or$-135°$ from positive x axis in the clock wise direction.

The position vectors for the next half hour are,

$\overrightarrow{r_0}=\left(-10\mathrm{cm}\right)\hat{\mathrm{j}}\mathrm{and}\overrightarrow{\mathrm{r}}=\left(10\mathrm{cm}\right)\hat{\mathrm{j}}$

So, the displacement vector for next half hour is,

$$\begin{gathered} ∆\overrightarrow{r}=\left(10\mathrm{cm}\right)\hat{\mathrm{j}}-\left(-10\mathrm{cm}\right)\hat{\mathrm{j}} \\ =\left(20\mathrm{cm}\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude for the next half hour is,

$\left|∆\overrightarrow{r}\right|=20\mathrm{cm}$

Thus, the magnitude of displacement for next half hour is 20 cm.

Using equation (iii), the direction of displacement is,

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{20\mathrm{cm}}{0}\right) \\ =90° \end{gathered}$$

Thus, the direction of displacement is${}^{90°}$from the positive x axis in the counter clockwise direction.

In the full hour sweep, the displacement becomes zero as the hand returns to its starting position.

Since, the displacement for a full hour sweep is zero therefore, the corresponding angle for the full hour sweep is also zero.