Problem 47 - College Physics

Question

A car on a circular track accelerates from rest. (a) The car experiences (1) only angular acceleration, (2) only centripetal acceleration, (3) both angular and centripetal accelerations. Why? (b) If the radius of the track is \(0.30 \mathrm{~km}\) and the magnitude of the constant angular acceleration is \(4.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2},\) how long does the car take to make one lap around the track? (c) What is the total (vector) acceleration of the car when it has completed half of a lap?

Step-by-Step Solution

Answer

(a) Both angular and centripetal accelerations; (b) 117.5 seconds; (c) 21.1 m/s².

Step 1: Analyze Accelerations

When a car starts accelerating on a circular track, it initially has only angular acceleration because it starts from rest and needs to build up its angular velocity. As it begins to move in a circle, it experiences centripetal acceleration to change the direction of the velocity. Thus, the car experiences both angular and centripetal accelerations.

Step 2: Determine the Time to Complete a Lap

First, we need to find the final angular velocity when the car completes a lap. Given the radius of the track is 0.3 km or 300 meters. The angular distance for one lap is \( 2\pi \) radians. Using the kinematic equation for angular motion \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), where \( \omega_0 \) is initial angular velocity, \( \alpha \) is angular acceleration, and \( t \) is time. Since \( \omega_0 = 0 \), the equation simplifies to \( \theta = \frac{1}{2} \alpha t^2 \). Solving for \( t \) when \( \theta = 2\pi \), we have \( 2\pi = \frac{1}{2} \times 4.5 \times 10^{-3} \times t^2 \). Solving for \( t \), \( t = \sqrt{\frac{2 \times 2\pi}{4.5 \times 10^{-3}}} \).

Step 3: Calculate the Time

Calculate \( t \) using the equation: \( t = \sqrt{\frac{4\pi}{4.5 \times 10^{-3}}} \). This results in \( t \approx 117.5 \) seconds.

Step 4: Finding Total Acceleration at Half Lap

At half a lap, \( \theta = \pi \). The angular velocity \( \omega \) can be found using \( \omega = \alpha t \). For half a lap, \( t' = t/2 \approx 58.75 \) seconds. Hence, \( \omega = 4.5 \times 10^{-3} \times 58.75 \approx 0.264 rad/s \). The centripetal acceleration \( a_c = \omega^2 \times r \). Substitute the values: \( a_c \approx 0.264^2 \times 300 \approx 21.05 \ m/s^2 \). The angular acceleration \( a_t = \alpha \times r = 4.5 \times 10^{-3} \times 300 = 1.35 \ m/s^2 \).

Step 5: Calculate Vector Acceleration

The total acceleration is a vector sum of tangential and centripetal accelerations. Use \( a = \sqrt{a_t^2 + a_c^2} \). Compute: \( a = \sqrt{1.35^2 + 21.05^2} \approx 21.1 \ m/s^2 \).

Key Concepts

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. It is a fundamental concept when describing rotational motion. If you've ever watched a spinning top start its spin, you've witnessed angular acceleration in action. In the context of the circular track problem, the car accelerates from rest, meaning it initially has no angular velocity. But as the engine works, the car's wheels begin turning, increasing its speed along the path. This change in angular velocity is what we call angular acceleration.

Key points about angular acceleration:

• Characterized by the symbol \( \alpha \).
• Measured in \( \text{rad/s}^2 \) (radians per second squared).
• Relates to linear acceleration through the radius \( r \) by the equation \( a_t = r \times \alpha \), where \( a_t \) is the tangential acceleration.

Angular acceleration allows us to understand how fast an object is picking up speed as it rotates around a point. In the problem, the car's angular acceleration is given as \( 4.5 \times 10^{-3} \text{rad/s}^2 \). This tells us how quickly the car ramps up its rotational speed as it traverses the circular path.

Centripetal Acceleration

Centripetal acceleration is crucial for any object traveling in a circular path. It is what keeps the car on the track and directs it towards the center of the circle. Unlike angular acceleration, which changes the speed of an object, centripetal acceleration changes the direction of the velocity vector, helping the object maintain a circular path.

Some important aspects of centripetal acceleration include:

• Symbolized as \( a_c \).
• Calculated using the formula \( a_c = \omega^2 \times r \), where \( \omega \) is the angular velocity.
• Measured in \( \text{m/s}^2 \) (meters per second squared).
• Provides the inward force necessary for circular motion.

In our car example, as the vehicle speeds up, its centripetal acceleration increases. This ensures the car can follow the curve of the track smoothly. For a car traveling at an angular velocity of \( 0.264 \text{rad/s} \) on a radius of 300 meters, the centripetal acceleration would be about \( 21.05 \text{m/s}^2 \).

Vector Acceleration

Vector acceleration is a combination of linear (tangential) and centripetal accelerations, taking vector direction into account. Since both of these accelerations are perpendicular, they can be combined using the Pythagorean theorem to find the total acceleration. This gives us a comprehensive understanding of the acceleration the car experiences.

When combining accelerations, keep in mind:

• Resultant acceleration is found using \( a = \sqrt{a_t^2 + a_c^2} \).
• \( a_t \) is the tangential (linear) acceleration.
• Vector diagrams can help visualize how these two components combine.

In the exercise, the vector sum of tangential and centripetal acceleration gives a total acceleration of roughly \( 21.1 \text{m/s}^2 \). This value represents the entire force the car feels as it moves halfway around the track.

Angular Kinematics

Angular kinematics is the study of motion with respect to rotation, similar to how linear kinematics applies to straight-line motion. It's essential for predicting how the position, velocity, and acceleration change over time for rotating objects. Just as linear kinematics uses position, velocity, and acceleration, angular kinematics uses angular displacement, velocity, and acceleration.

Key formulas in angular kinematics include:

• \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) (angular displacement).
• \( \omega = \omega_0 + \alpha t \) (angular velocity).
• Similar to linear kinematics, but with rotational terms.

For our car, starting from rest means initial angular velocity \( \omega_0 = 0 \). The car's angular position after time \( t \) can be found using the displacement formula. This tells us how far the car has rotated on the track. Understanding angular kinematics allows us to predict the car's position and speed at any time during its motion.