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As your bus rounds a flat curve at constant speed of \(20 \mathrm{~m} / \mathrm{s}\) a package with mass \(0.500 \mathrm{~kg}\), suspended from the luggage compartment of the bus by a string \(45.0 \mathrm{~cm}\) long, is found to hang at rest relative to the bus, with the string making an angle of \(30.0^{8}\) with the vertical. (a) What is the radial acceleration of the bus? (b) What is the radius of the curve?

Short Answer

Expert verified
(a) Radial acceleration is approximately 5.66 m/s². (b) Radius of the curve is approximately 70.74 m.

Step by step solution

01

Understand the Forces

The package is in equilibrium while the bus moves through the curve. The forces acting on it are tension in the string, gravitational force, and centripetal force due to the bus's motion. The tension has a component balancing gravity (vertical) and a component providing centripetal force (horizontal).
02

Resolve Forces Vertically

Vertically, the component of the tension that balances the weight of the package is given by the equation:\[ T \cos(\theta) = mg \]where \( T \) is the tension, \( \theta = 30^\circ \), \( m = 0.500 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \).
03

Resolve Forces Horizontally

Horizontally, the component of tension that provides the centripetal force is:\[ T \sin(\theta) = ma_r \]where \( a_r \) is the radial acceleration. Combine this with the vertical component to derive expressions involving \( T \).
04

Calculate Radial Acceleration

Combine the vertical and horizontal components to eliminate \( T \):\[ \frac{T \sin(\theta)}{T \cos(\theta)} = \frac{a_r}{g} \ \tan(\theta) = \frac{a_r}{g} \a_r = g \tan(\theta) \]Plug in the values to find:\[ a_r = 9.8 \, \text{m/s}^2 \times \tan(30^\circ) \approx 5.66 \, \text{m/s}^2 \]
05

Use Centripetal Formula

The centripetal acceleration is also defined by \( a_r = \frac{v^2}{r} \), where \( v = 20 \, \text{m/s} \). Knowing \( a_r = 5.66 \, \text{m/s}^2 \), we solve for \( r \) (radius of the curve).
06

Calculate Radius of the Curve

Rearrange the centripetal formula to solve for \( r \):\[ r = \frac{v^2}{a_r} \r = \frac{(20 \, \text{m/s})^2}{5.66 \, \text{m/s}^2} \approx 70.74 \, \text{m} \]

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Key Concepts

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

Radial Acceleration
Radial acceleration comes into play in situations involving circular motion. It occurs because an object moving in a circle is always changing direction. Think of a roller coaster ride. As the coaster car turns, passengers feel a force pushing them sideways. That's your radial acceleration at work! In physics, radial acceleration is directed towards the center of the circular path. This is why it is sometimes called "centripetal acceleration"—with "centripetal" literally meaning "center seeking." Let's break this down a bit further.

Consider any object in circular motion: the change in velocity always points towards the center. This change in velocity means the object is accelerating, even if its speed is constant. Thus, radial acceleration is key to keeping the object on its curved path. In the bus exercise, calculating radial acceleration helps determine the forces at work. You'll need to use the formula:

\[ a_r = g \tan(\theta) \]
where \( g \) is gravity and \( \theta \) is the angle given.
  • It's all about understanding how motion in a curve depends on both speed and the curve's size.
  • Radial acceleration is not just about speed, but how rapidly that speed direction changes.
Circular Motion
Circular motion is everywhere—from planets orbiting stars to children spinning on merry-go-rounds. At its core, circular motion arises when an object travels along a circular path. However, for an object to maintain this path, it must constantly experience a force pulling it towards the center of the circle. This force is known as the centripetal force.

In the case of our bus problem, the package inside experiences circular motion as the bus rounds the curve. Two forces are at play here: gravitational force pulling downward, and the tension from the string pulling both up and towards the bus's center of motion. For circular motion, it is essential to resolve these forces to understand the motion's dynamics.

  • Circular motion examples include a spinning wheel, electrons in an atom, and even Earth's orbit around the sun.
  • Understanding circular motion helps in designing safe roads, roller coasters, and vehicle dynamics.

Remember, in circular motion, speed might be constant, but velocity (a vector) changes due to direction shifts.
Physics Problems
Physics problems, like the bus and package example, are vital for applying theoretical knowledge to real-world situations. Solving physics problems often requires dissecting complex scenarios into manageable parts, like breaking down the forces in our circular motion example.

The primary goal is to identify knowns and unknowns—like the speed of the bus or the angle of the string—and apply physics principles accordingly. In our exercise:
  • Forces are resolved into vertical and horizontal components to simplify the problem.
  • Equations like \[ T \cos(\theta) = mg \] and \[ T \sin(\theta) = ma_r \] help isolate variables.
By solving these problems, students gain a deeper understanding of physics concepts and improve critical thinking skills. Approaching problems with patience and a step-by-step methodology is often the key to success.
Physics Education
Physics education provides the tools to explore and understand the natural laws governing our world. It emphasizes clear thinking and problem-solving skills. The example of calculating the radial acceleration and the curve's radius illustrates how physics education helps bridge theory with practical applications.

One of the best ways to learn in physics is through problem-solving exercises. They guide students to:
  • Grasp fundamental concepts, such as understanding motion dynamics.
  • Decipher complex situations into simpler parts, like resolving forces.
  • Apply mathematical formulas to solve real-world problems.
By incorporating exercises similar to the one discussed, physics education encourages exploration and curiosity. As students tackle more physics problems, they become more intuitive in predicting outcomes and recognizing real-world applications.

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Most popular questions from this chapter

A highway curve with radius \(900.0 \mathrm{ft}\) is to be banked so that a car traveling 55.0 mph will not skid sideways even in the absence of friction. (a) Make a free-body diagram of this car. (b) At what angle should the curve be banked?

Consider the fact that an object sitting at the equator is undergoing uniform circular motion due to the rotation of the earth. (a) Draw a free-body diagram for a box of mass \(m\) that is placed at the equator. (b) Now suppose that we allow the day to become shorter than 24 h. Use Newton's second law to calculate the maximum speed, due to the earth's rotation, that the box can have and still remain on the surface. (c) How long would this day be in hours?

The star Rho' Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho \({ }^{1}\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho \(^{1}\) Cancri?

The asteroid 243 Ida has a mass of about \(4.0 \times 10^{16} \mathrm{~kg}\) and an average radius of about \(16 \mathrm{~km}\) (it's not spherical, but you can assume it is). (a) Calculate the acceleration of gravity on 243 Ida. (b) What would an astronaut whose earth weight is \(650 \mathrm{~N}\) weigh on 243 Ida? (c) If you dropped a rock from a height of \(1.0 \mathrm{~m}\) on 243 Ida, how long would it take for the rock to reach the ground? (d) If you can jump \(60 \mathrm{~cm}\) straight up on earth, how high could you jump on 243 Ida? (Assume the asteroid's gravity doesn't weaken significantly over the distance of your jump.)

A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m}\). Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of her hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

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