Problem 4 - Fundamentals of Physics

Question

The minute hand of a wall clock measures \(10 \mathrm{~cm}\) 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?

Step-by-Step Solution

Answer

(a) 14.14 cm, (b) 90°, (c) 20 cm, (d) 180°, (e) 0 cm, (f) 0°.

Step 1: Understanding the Problem

We need to calculate the displacement vector of the tip of the clock's minute hand for three different time intervals. This involves determining the difference in position of the tip of the minute hand at two different times. We'll break it down into three parts corresponding to each time interval.

Step 2: Calculating the Magnitude of Displacement for Quarter to Half Past the Hour

In the first interval from a quarter past the hour (15 minutes) to half past the hour (30 minutes), the minute hand covers \(\frac{1}{4}\) of the complete circle (from 15 minutes to 30 minutes, which is 15 minutes of a 60-minute hour). The formula for displacement magnitude in a circle is \ |\overrightarrow{D}| = 2r \sin\left(\frac{\theta}{2}\right) \, where \ r = 10 \,\mathrm{cm} \ and \ \theta = 90^\circ \, (quarter of 360 degrees). \ Thus, \ |\overrightarrow{D}| = 2 \times 10 \times \sin(45^\circ) = 20 \times \frac{\sqrt{2}}{2} = 10 \sqrt{2} \, approximately equal to 14.14 cm.

Step 3: Calculating the Angle for Quarter to Half Past the Hour

The angle of the displacement vector with respect to the starting point is the same as the angle covered by the minute hand over this period, which is 90 degrees.

Step 4: Calculating the Magnitude of Displacement for the Next Half Hour

From half past the hour (30 minutes) to top of the hour (60 minutes), the minute hand covers another \(\frac{1}{2}\) of the circle (180 degrees). Using the formula \ |\overrightarrow{D}| = 2r \sin\left(\frac{\theta}{2}\right) \, where \ \theta = 180^\circ \, we get \ |\overrightarrow{D}| = 2 \times 10 \times \sin(90^\circ) = 20 \times 1 = 20 \,. Thus, the displacement magnitude for the next half hour is 20 cm.

Step 5: Calculating the Angle for the Next Half Hour

Here, the angle of the displacement vector relative to the starting point is 180 degrees, as the minute hand moves across half the circle.

Step 6: Calculating the Magnitude of Displacement for the Next Hour

In the next hour, the minute hand makes a full circle, meaning the initial and final positions are the same. Thus, the magnitude of displacement is 0 cm.

Step 7: Calculating the Angle for the Next Hour

Since the starting and ending points of this movement are the same (completing a full circle), the angle is not defined as there is no net displacement, which results in an angle of 0 degrees.

Key Concepts

Minutes to Degrees Conversion

Understanding how to convert minutes into degrees is crucial when dealing with problems involving circular motion, like the movement of the minute hand on a clock. A full circle is composed of 360 degrees, and since there are 60 minutes in an hour, each minute represents a specific fraction of the circle. By dividing 360 degrees by 60 minutes, you find that each minute corresponds to 6 degrees of rotation.

Here's a quick way to make these conversions:

• 1 minute = 6 degrees
• 15 minutes = 15 × 6 = 90 degrees
• 30 minutes = 30 × 6 = 180 degrees
• 45 minutes = 45 × 6 = 270 degrees
• 60 minutes = 60 × 6 = 360 degrees

This conversion helps simplify calculations and provides a better understanding of the clock's minute hand movements. By converting minutes to degrees, students can accurately determine angles and analyze problems involving circular motion.

Circular Motion

Circular motion describes the movement of an object along a circular path. In physics, it is important to understand how different parameters describe this motion, such as the radius of the circle, the angular displacement, and the period of rotation.

The radius, in the case of the clock, is the length from the center of the clock face to the tip of the minute hand. In our example, this is given as 10 cm.

• The **angular displacement** is the angle through which the minute hand moves. It is expressed in degrees or radians.
• A **complete circle** represents an angular displacement of 360 degrees or 2π radians.
• The **period** is the time it takes for the minute hand to complete one full rotation, which is 60 minutes for a standard clock.
  
  

For our specific problem, in every part of the clock's motion, we describe the displacement in terms of the part of the circle traversed during a specific time interval. Understanding these fundamental aspects of circular motion helps solve many physics problems involving both rotation and periodic motion.

Trigonometric Functions in Physics

Trigonometric functions play a significant role in physics when analyzing the characteristics of circular motion and determining vector components. These functions, such as sine and cosine, help us calculate magnitudes and angles concerning displacement vectors.

For the exercise, the displacement of the minute hand is calculated using the formula: \[\left| \overrightarrow{D} \right| = 2r \sin\left(\frac{\theta}{2}\right)\] Where:

• \( \left| \overrightarrow{D} \right| \) is the magnitude of the displacement vector,
• \( r \) is the radius (or length of the minute hand),
• \( \theta \) is the angle, in degrees, covered by the minute hand during the given time interval.

The sine function is particularly useful for determining the magnitude because it helps describe how far "out" the hand has moved from its initial position in a straight line. The vector analysis is crucial to synthesize both the direction and the magnitude of displacement, allowing students to understand how trigonometric functions aid in depicting complex physical relationships in rotational motion.