Problem 52 - College Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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).

Step 2: 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 \( g = 9.8 \, \text{m/s}^2 \).

Step 3: 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 \( T \).

Step 4: 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:\[ a_r = 9.8 \, \text{m/s}^2 \times \tan(30^\circ) \approx 5.66 \, \text{m/s}^2 \]

Step 5: 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).

Step 6: Calculate Radius of the Curve

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

Key Concepts

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.