Problem 104 - University Physics with Modern Physics

Question

You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with mass \(0.500 \mathrm{kg},\) suspended from the ceiling of the bus by a string 1.80 \(\mathrm{m}\) long, is found to hang at rest relative to the bus when the string makes an angle of \(30.0^{\circ}\) with the vertical. In this position the lunch box is 50.0 \(\mathrm{m}\) from the center of curvature of the curve. What is the speed \(v\) of the bus?

Step-by-Step Solution

Answer

The speed of the bus is approximately 9.5 m/s.

Step 1: Analyze Forces

When the bus rounds the curve, the lunch box is subject to two forces: the gravitational force vertical downwards and the tension in the string, which can be divided into components. The gravitational force is given by \( mg \) where \( m = 0.5 \) kg and \( g = 9.8 \) m/s². The tension force can be split into two components: the vertical component, \( T\cos(\theta) \), which balances the gravitational force, and the horizontal component, \( T\sin(\theta) \), which provides the necessary centripetal force for circular motion.

Step 2: Set Up Equations

From the balance of forces, we know: 1. For vertical forces: \( T\cos(\theta) = mg \). 2. For horizontal forces providing centripetal force: \( T\sin(\theta) = \frac{mv^2}{r} \) where \( r \) is the radius of curvature of the curve (50.0 m).

Step 3: Solve for Tension (T)

We can solve the first equation for tension \( T \): \[ T = \frac{mg}{\cos(\theta)} \] Substituting in values: \( T = \frac{0.5 \times 9.8}{\cos(30^\circ)} \approx 5.66 \) N.

Step 4: Solve for Speed (v)

Next, we solve the centripetal force equation for \( v \). Substitute \( T \) from the previous step into the second equation: \[ \frac{mg\sin(\theta)}{\cos(\theta)} = \frac{mv^2}{r} \]Simplify this equation to find \( v \): \[ v^2 = rg \frac{\sin(\theta)}{\cos(\theta)} \]Thus, solving for \( v \): \[ v = \sqrt{rg\tan(\theta)} \]Substitute \( r = 50 \) m, \( g = 9.8 \) m/s², and \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[ v = \sqrt{50 \times 9.8 \times \frac{1}{\sqrt{3}}} \approx 9.5 \text{ m/s} \]

Step 5: Conclusion: Calculate Speed

The bus travels at a speed of approximately 9.5 m/s when making the turn while the lunch box remains at an angle.

Key Concepts

Centripetal Force

When an object moves in a circular path, even at a constant speed, it experiences centripetal force. This force acts towards the center of the circle and is necessary for keeping the object in its curved path. In the case of the lunch box in the bus, the centripetal force is provided by the horizontal component of the tension in the string. The tension component must pull towards the center of the circle. This ensures the lunch box follows the bus' circular motion, rather than flying off in a straight line due to inertia.

The mathematical expression for this force is given by:

• \( F_c = \frac{mv^2}{r} \)
• Where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the circle.

Understanding centripetal force helps us model and predict the motion of objects in various rotating systems, from simple pendulum swings to complex planetary orbits.

Tension in String

Tension in a string acts as a pulling force that can have components in multiple directions based on how the string is oriented. In our example with the lunch box, the tension force has two components:

• Vertical Component: \( T\cos(\theta) \) which counteracts the gravitational force.
• Horizontal Component: \( T\sin(\theta) \) that acts as the centripetal force.

These components are crucial because they illustrate how forces balance each other out. With the vertical balance, the lunch box doesn't rise or fall, affirming its equilibrium state. The horizontal component allows the box to "feel" the turn but stay taut to the line of motion.

To find the tension, you leverage trigonometry and known values. With the formula:

• \( T = \frac{mg}{\cos(\theta)} \)

This formula allows us to solve for tension (\( T \)) by rearranging based on vertical force equilibrium. Calculating tension in these scenarios helps in understanding force magnitudes and directions in physical systems.

Trigonometry in Physics

Trigonometry is key in physics, offering tools to decompose and analyze forces acting at angles. In circular motion, trigonometry allows us to split forces into perpendicular components. This is exactly what happens with our lunch box scenario where the string makes an angle with the vertical.

Key trigonometric functions, such as sine, cosine, and tangent, become important:

• \( \cos(\theta) \) is used for the vertical component of tension because it aligns with the force of gravity.
• \( \sin(\theta) \) helps find the horizontal component, crucial for the centripetal force.
• \( \tan(\theta) \) is used when simplifying the relationship between forces. This is evident in the equation \( v = \sqrt{rg\tan(\theta)} \).

Applying these functions allows us to break down complex physics problems into simpler, solvable components. Trigonometry is immensely valuable for any motion analysis.

Gravitational Force Analysis

Gravitational force, a fundamental physics concept, acts downwards, influencing various aspects of our scenario. Given by \( F_g = mg \), this force is constant in scenarios close to the surface of the Earth.

In the example with the school bus and lunch box, gravitational force plays a key role by defining the weight of the lunchbox. It ensures there is a component of tension in the string counteracting this force, keeping the box stationary except for its swing out to the side.

• The vertical component of tension \( T\cos(\theta) \) equals \( mg \), keeps gravitational forces in check.

A proper understanding of gravitational force analysis allows students to predict behaviors in real-world and theoretical physics problems by establishing early equations which simplify the problem.