Problem 55 - College Physics

Question

As your bus rounds a flat curve at constant speed, 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 \(30.0^{\circ}\) with the vertical. In this position, the package is 50.0 \(\mathrm{m}\) from the center of curvature of the curve. What is the speed of the bus?

Step-by-Step Solution

Answer

The speed of the bus is approximately 12.8 m/s.

Step 1: Understand the Scenario

We have a bus taking a curve and a package hanging inside the bus. The string suspending the package makes an angle of 30 degrees with the vertical, indicating a centripetal force acting on the package due to the curve.

Step 2: Identify the Forces

The forces acting on the package are tension in the string and gravity. The tension has two components: one vertical (balancing gravity) and one horizontal (providing the centripetal force).

Step 3: Resolve the Tension into Components

The tension (T) can be resolved into:- Vertical component: \(T_y = T \cos(30^{\circ})\)- Horizontal component providing centripetal force: \(T_x = T \sin(30^{\circ})\)

Step 4: Apply Newton's Second Law Vertically

Since the package is at rest vertically: \\[ T \cos(30^{\circ}) = mg \] \where \(m = 0.500\, \text{kg}\).

Step 5: Apply Newton's Second Law Horizontally

The horizontal component of tension provides the centripetal force needed for circular motion: \\[ T \sin(30^{\circ}) = \frac{m v^2}{r} \] \where \(r = 50.0\, \text{m}\).

Step 6: Combine Equations

From the vertical force balance, \\[ T = \frac{mg}{\cos(30^{\circ})} \] \Substitute into the horizontal force equation to find \(v\): \\[ \frac{mg \sin(30^{\circ})}{\cos(30^{\circ})} = \frac{m v^2}{r} \]

Step 7: Solve for the Speed

Cancel \(m\) and solve for \(v\): \[ v^2 = rg \tan(30^{\circ}) \] \[ v = \sqrt{50.0 \times 9.81 \times \frac{1}{\sqrt{3}}} \]Calculating, \v \approx 12.8 \, \text{m/s}.

Key Concepts

Centripetal Force

Centripetal force is the force that acts on an object moving in a circular path and is directed towards the center around which the object is moving. It is essential for keeping the object in motion along the circular path. When a package is hanging inside a bus moving around a curve, it experiences this centripetal force.
In the given scenario, the centripetal force is not an additional force but rather comes from the horizontal component of the tension in the string. Essentially:

• The object (package) tends to move in a straight line due to inertia.
• The tension in the string provides the needed inward or centripetal force to keep it moving in a circular path.
• This force ensures that the package stays in its circular trajectory parallel to the curve taken by the bus.

Understanding centripetal force helps reveal why the package hangs at a specific angle in relation to the vertical while the bus navigates the curve.

Newton's Second Law

Newton's Second Law states that the force acting on an object is equal to the mass of that object times its acceleration (\[ F = ma \]). It's crucial for analyzing the forces at play in any physics problem involving motion. In the context of circular motion:

• The acceleration is directed towards the center of the circular path (centripetal acceleration).
• The centripetal force can be expressed as \[ F_c = rac{mv^2}{r} \], where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius of the curve.
• This equation clarifies how changes in speed or radius affect the required centripetal force.

Newton's Second Law is key to establishing the relationship between the forces we observe—tension, gravity—and the centripetal force required for circular motion. By resolving the forces in both vertical and horizontal components, it helps further understand how the package behaves in the scenario.

Trigonometry in Physics

Trigonometry is often used in physics to resolve forces into their components. In this exercise, the angle between the string and the vertical direction calls for trigonometric functions:

• Cosine and sine functions help break down the tension into vertical and horizontal components.
• The vertical component \( T_y = T \cos(30^{\circ}) \) balances the weight of the package, \( mg \).
• The horizontal component \( T_x = T \sin(30^{\circ}) \) supplies the centripetal force needed for the circular motion.

Trigonometry allows us to solve for unknowns indirectly by relating angles and side lengths. The use of \[\tan(30^{\circ})\] was crucial for finding the bus's speed by linking it to the radius and gravitational force. This mathematical approach simplifies complex problems by breaking down forces into simpler, calculable parts.