Problem 2 - Applied Mathematics

Question

A particle is describing a circle of radius \(4 \mathrm{~m}\) with a constant angular accel. eration. At one instant it has a speed of \(2 \mathrm{~ms}^{-1}\) and \(4 \mathrm{~s}\) later it has a speed of \(10 \mathrm{~ms}^{-1}\). Find its angular acceleration and the distance it has travelled in this time.

Step-by-Step Solution

Answer

Angular acceleration is \(0.5 \text{ rad/s}^2\) and distance travelled is \( 24 \text{ m} \).

Step 1: Express initial and final speeds in terms of angular velocity

The speed of a particle in circular motion is given by the product of the radius and the angular velocity: \( v = r \theta \). Given the initial speed \( v_0 = 2 \text{ m/s} \) and the final speed \( v = 10 \text{ m/s} \), we can write \( 2 = 4 \theta_0 \) and \( 10 = 4 \theta \). Solving for \( \theta_0 \) and \( \theta \), we get \( \theta_0 = 0.5 \text{ rad/s} \) and \( \theta = 2.5 \text{ rad/s} \).

Step 2: Calculate the angular acceleration

We know the time interval \( \triangle t = 4 \text{ s} \), and the change in angular velocity \( \triangle \theta = \theta - \theta_0 = 2.5 - 0.5 = 2 \text{ rad/s} \). Using the formula for angular acceleration \( \beta = \frac{\triangle \theta}{\triangle t} \), we get \( \beta = \frac{2}{4} = 0.5 \text{ rad/s}^2 \).

Step 3: Find the distance travelled

The angular displacement \( \theta(t) \) under constant angular acceleration can be calculated using \( \theta(t) = \theta_0 t + \frac{1}{2}\beta t^2 \). Substituting \( \theta_0 = 0.5 \text{ rad/s} \), \( \beta = 0.5 \text{ rad/s}^2 \), and \( t = 4 \text{ s} \), we get \( \theta(4) = 0.5 \times 4 + \frac{1}{2} \times 0.5 \times 4^2 = 2 + 4 = 6 \text{ rad} \). The linear distance travelled is \( s = r \theta \), so \( s = 4 \times 6 = 24 \text{ m} \).

Key Concepts

Angular Acceleration

Angular acceleration is the rate at which the angular velocity of an object changes with time. It is denoted by \( \beta \) and is measured in radians per second squared (\( \text{rad/s}^2 \)). In this problem, we determined the angular acceleration using the formula: \[ \beta = \frac{\Delta \theta}{\Delta t} \] Here, \( \Delta \theta \) is the change in angular velocity, and \( \Delta t \) is the time over which this change occurs. Given the initial angular velocity (\( \theta_0 = 0.5 \text{ rad/s} \)) and the final angular velocity (\( \theta = 2.5 \text{ rad/s} \)), and the time interval of 4 seconds, we found: \[ \beta = \frac{2.5 - 0.5}{4} = 0.5 \text{ rad/s}^2 \] This shows that the particle's angular velocity increases by 0.5 radians per second every second.

Angular Velocity

Angular velocity, denoted by \( \theta \), describes how quickly an object rotates or revolves relative to another point. It is typically measured in radians per second (\( \text{rad/s} \)). In circular motion, the linear speed \( v \) of an object is related to its angular velocity by the radius of the circle: \[ v = r \theta \] For the given problem, the radius is 4 meters. The particle's initial speed was 2 m/s, and 4 seconds later, its speed was 10 m/s. Converting these speeds into angular velocities, we have: \[ \theta_0 = \frac{2}{4} = 0.5 \text{ rad/s} \] and \[ \theta = \frac{10}{4} = 2.5 \text{ rad/s} \] These values indicate how fast the particle is spinning at the start and end of the observed time period.

Linear Distance

Linear distance traveled in circular motion is the actual path length covered along the circumference of the circle. It can be calculated using the angular displacement and the radius of the circle: \[ s = r \theta \] For angular displacement under constant angular acceleration, we use: \[ \theta(t) = \theta_0 t + \frac{1}{2} \beta t^2 \] Here, \( \theta_0 \) is the initial angular velocity, \( \beta \) is the angular acceleration, and \( t \) is the time. For our problem, substituting \( \theta_0 = 0.5 \text{ rad/s} \), \( \beta = 0.5 \text{ rad/s}^2 \), and \( t = 4 \text{ s} \), we get: \[ \theta(4) = 0.5 \times 4 + \frac{1}{2} \times 0.5 \times 4^2 = 2 + 4 = 6 \text{ rad} \] The total linear distance covered is then: \[ s = 4 \times 6 = 24 \text{ m} \] Thus, the particle travels a linear distance of 24 meters over the 4-second period.