Problem 14 - Applied Physics

Question

A cycle of mass \(510 \mathrm{~kg}\) rounds a curve of radius \(4 \overline{0} \mathrm{~m}\) at \(95 \mathrm{~km} / \mathrm{h}\). What is the centripetal force on the cycle?

Step-by-Step Solution

Answer

The centripetal force on the cycle is approximately 8880 N.

Step 1: Convert Speed to Meters per Second

First, we need to convert the speed from kilometers per hour (km/h) to meters per second (m/s). The speed is given as 95 km/h. To convert, use the conversion factor: \(1~\text{km/h} = \frac{1000}{3600}~\text{m/s}\). Thus, \(95~\text{km/h} = 95 \times \frac{1000}{3600}~\text{m/s} = \frac{95 \times 1000}{3600}~\text{m/s} \approx 26.39~\text{m/s}\).

Step 2: Use Centripetal Force Formula

The formula for centripetal force \( F_c \) is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the curve. Here, \( m = 510~\text{kg} \), \( v = 26.39~\text{m/s} \), and \( r = 40~\text{m} \).

Step 3: Compute Velocity Squared

Calculate \( v^2 \) by squaring the velocity: \( v^2 = (26.39)^2 \approx 696.47~\text{m}^2/\text{s}^2 \).

Step 4: Calculate Centripetal Force

Substitute the values into the centripetal force formula: \( F_c = \frac{510 \times 696.47}{40} \). This simplifies to \( F_c = \frac{355199.7}{40} \approx 8879.99~\text{N} \).

Step 5: Round the Final Answer

Round the computed centripetal force to a reasonable number of significant figures. Thus, the centripetal force is \( F_c \approx 8880~\text{N} \).

Key Concepts

Conversions in Physics

Physics frequently requires us to convert units in order to properly solve problems. Conversions allow us to communicate measurements in a consistent way. In the context of physics, common units may need to be converted to those of the International System of Units (SI) for consistency and ease of calculation. For instance, speed is often given in kilometers per hour (km/h), but calculations in physics typically use meters per second (m/s). To convert speed:

• Recognize that 1 km = 1000 meters and 1 hour = 3600 seconds.
• Thus, 1 ext{ km/h} = \( \frac{1000}{3600} \text{ m/s} \approx 0.2778 \text{ m/s} \).

This factor is crucial when ensuring accuracy in our computations as seen in the speed conversion in the exercise.

Speed Calculation

The calculation of speed and understanding velocity are foundational in physics problems involving motion. Speed is the rate at which an object covers distance. In this case, speed is given, but it's crucial to calculate or convert it accurately to apply formulas based on motion in physics. Here, the speed of the cycle is initially 95 km/h. After conversion to meters per second, it becomes 26.39 m/s. Knowing the correct speed in the necessary units is fundamental for problems involving forces like centripetal force.

Centripetal Force Formula

Centripetal force is the force required to make an object follow a curved path. The formula to calculate centripetal force \( F_c \) is crucial in solving problems involving circular motion. This formula is: \[ F_c = \frac{mv^2}{r} \] where:

• \( m \) is the mass in kilograms.
• \( v \) is the velocity in meters per second.
• \( r \) is the radius of the path in meters.

The force depends directly on the mass, the square of the velocity, and inversely on the radius. Understand this relationship helps in analyzing motion under the influence of centripetal force.

Problem-Solving in Physics

Problem-solving in physics often involves applying mathematical formulas to theoretical scenarios. The process involves:

• Understanding the given problem and identifying what needs to be calculated.
• Converting units to ensure all measurements align with the required formula units.
• Applying relevant physics formulas, like the centripetal force formula in this case.
• Solving step-by-step, keeping track of calculations and intermediate values.
• Rounding the final answer appropriately to reflect significant figures and improve readability.

These strategies provide structure, allowing you to focus on understanding the principles behind each physics challenge clearly, which is pivotal to successfully solving homework exercises.