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Chapter 5: Problem 104

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?

Short Answer

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

Step by step solution

01

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/s虏. 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.
02

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).
03

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.
04

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/s虏, and \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \): \[ v = \sqrt{50 \times 9.8 \times \frac{1}{\sqrt{3}}} \approx 9.5 \text{ m/s} \]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

Block \(A,\) with weight \(3 w,\) slides down an inclined plane \(S\) of slope angle \(36.9^{\circ}\) at a constant speed while plank \(B\) with weight \(w,\) rests on top of A. The plank is attached by a cord to the wall (Fig. P5.99). (a) Draw a diagram of all the forces acting on block A. (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A,\) determine its value.

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a second hanging block with mass \(m_{2}\) by a cord passing over a small, frictionless pulley (Fig. P5.68). The coefficient of static friction is \(\mu_{\mathrm{s}}\) and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\) (a) Find the mass \(m_{2}\) for which block \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the mass \(m_{2}\) for which block \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

An astronaut is inside a \(2.25 \times 10^{6} \mathrm{kg}\) rocket that is blasting off vertically from the launch pad. You want this rocket to reach the speed of sound \((331 \mathrm{m} / \mathrm{s})\) as quickly as possible, but you also do not want the astronaut to black out. Medical tests have shown that astronauts are in danger of blacking out at an acceleration greater than 4\(g .\) (a) What is the maximum thrust the engines of the rocket can have to just barely avoid blackout? Start with a free-body diagram of the rocket. (b) What force, in terms of her weight \(w,\) does the rocket exert on the astronatt? Start with a free-body diagram of the astronaut. (c) What is the shortest time it can take the rocket to reach the speed of sound?

You are riding your motorcycle one day down a wet street that slopes downward at an angle of \(20^{\circ}\) below the horizontal. As you start to ride down the hill, you notice a construction crew has dug a deep hole in the street at the bottom of the hill. A siberian tiger, escaped from the City Zoo, has taken up residence in the hole. You apply the brakes and lock your wheels at the top of the hill, where you are moving with a speed of 20 \(\mathrm{m} / \mathrm{s} .\) The inclined street in front of you is 40 \(\mathrm{m}\) long. (a) Will you plunge into the hole and become the tiger's lunch, or do you skid to a stop before you reach the hole? (The coefficients of friction between your motorcycle tires and the wet pavement are \(\mu_{\mathrm{s}}=0.90\) and \(\mu_{\mathrm{k}}=0.70 .\) ) ( b ) What must your initial speed be if you are to stop just before reaching the hole?
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