Question

4 The minute hand of a wall clock measures 12 cm from its tip to
the axis about which it rotates. The magnitude and angle of the dis-
placement vector of the tip are to be determined for three time inter-
vals. 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?​

Answer 1

Given that,

The minute hand of a wall clock measures 12 cm from its tip to the axis about which it rotates.

The vector distance is

$\vec{r_{1}}=(12\ cm) i$

$\vec{r_{2}}=(-12\ cm) j$

(a). We need to calculate the displacement vector

Using formula of displacement

$\Delta \vec{r}=\vec{r_{2}}-\vec{r_{1}}$

Put the value into the formula

$\Delta \vec{r}=(-12)i+(-12)j$

We need to calculate the magnitude of the displacement vector

Using formula of displacement

$|\Delta \vec{r}|=\sqrt{(-12)^2+(-12)^2}$

$|\Delta \vec{r}|=16.9\ cm$

We need to calculate the angle

Using formula of angle

$\theta=\tan^{-1}(\dfrac{j}{i})$

$\theta=\tan^{-1}(\dfrac{-12}{-12})$

$\theta=45^{\circ}$

(b). Now, the vector distance is

$\vec{r_{1}}=(-12\ cm) j$

$\vec{r_{2}}=(12\ cm) j$

We need to calculate the magnitude of the displacement

Using formula of magnitude of displacement

$|\Delta\vec{r}|=\vec{r_{2}}-\vec{r_{1}}$

Put the value into the formula

$|\Delta\vec{r}|=12j+12j$

$|\Delta\vec{r}|=24\ j$

We need to calculate the angle

Using formula of angle

$\theta=\tan^{-1}(\dfrac{j}{i})$

Put the value in to the formula

$\theta=\tan^{-1}(\dfrac{24}{0})$

$\theta=90^{\circ}$

(c). We need to calculate the magnitude of the displacement

The hand reaches at initial position in full hour

So, The displacement will be zero and angle also zero.

Hence, (a). The magnitude of displacement is 16.9 cm.

Angle is 45°

(b). The magnitude of displacement is 20 cm.

Angle is 90°

(c). The magnitude of displacement is zero.

Angle is zero.