Calculating Kinetic Energy of a Rotating Wheel

In this article, we will explore the calculation of kinetic energy for a rotating wheel. We will apply the principles of rotational motion to find the kinetic energy gained after 20 revolutions, starting from rest. We will use the given torque, moment of inertia, and angular displacement to determine the required parameters.

Overview of Rotational Motion

The equations of motion for constant linear acceleration can also be applied to rotational motion if the correct units are used. Let's start by understanding the basics and formulas required for our calculations.

Angular Acceleration

The angular acceleration ((alpha)) can be calculated using the torque ((T)) and the moment of inertia ((I)). The formula is:

[alpha frac{T}{I}]

Given the torque (T 200 , mathrm{Nm}) and the moment of inertia (I 100 , mathrm{kg cdot m^2}), we can find the angular acceleration as follows:

[alpha frac{200 , mathrm{Nm}}{100 , mathrm{kg cdot m^2}} 2 , mathrm{rad/s^2}]

Velocities and Angular Displacement

Next, we will determine the velocity ((omega)) and the angular displacement after 20 revolutions, starting from rest.

Velocity Calculation

The velocity after 4 seconds can be found using the kinematic equation:

[omega omega_i alpha t]

Given that the initial velocity (omega_i 0) (since it starts from rest), and (t 4 , mathrm{s}), we get:

[omega 0 2 , mathrm{rad/s^2} times 4 , mathrm{s} 8 , mathrm{rad/s}]

Angular Displacement Calculation

The angular displacement ((theta)) after 20 revolutions can be calculated as:

[theta 2pi times 20 , mathrm{revolutions} 40pi , mathrm{radians}]

Kinetic Energy Calculation

Finally, we can calculate the kinetic energy gained by the wheel after 20 revolutions. The formula for kinetic energy ((KE)) is:

[KE frac{1}{2} I omega^2]

Substituting the values:

[omega 20pi , mathrm{rad/s}] [I 100 , mathrm{kg cdot m^2}] [KE frac{1}{2} times 100 , mathrm{kg cdot m^2} times (20pi , mathrm{rad/s})^2 20000pi , mathrm{Joules}]

Therefore, the kinetic energy gained by the wheel after 20 revolutions starting from rest is (20000pi , mathrm{Joules}).

Additional Calculations

Let's verify the kinetic energy calculation using the equations for linear and rotational motion.

Linear Motion Equations

[S V_i t frac{1}{2} a t^2] [V_f V_i a t] [V_f^2 - V_i^2 2 a S]

Rotational Motion Equations

[theta omega_i t frac{1}{2} alpha t^2] [omega_f omega_i alpha t] [omega_f^2 - omega_i^2 2 alpha theta]

Using the torque and moment of inertia, we find the angular acceleration:

[alpha frac{T}{I} frac{200 , mathrm{Nm}}{100 , mathrm{kg cdot m^2}} 2 , mathrm{rad/s^2}]

Convert 20 revolutions to radians:

[40pi , mathrm{radians}]

Find the final angular velocity:

[omega_f^2 2 alpha theta 2 times 2 times 40pi 160pi , mathrm{rad^2/s^2}]

Calculate the kinetic energy:

[KE frac{1}{2} I omega_f^2 frac{1}{2} times 100 times 160pi 8000pi , mathrm{Joules}]

This confirms our previous calculations. Hence, the kinetic energy gained after 20 revolutions starting from rest is (8000pi , mathrm{Joules}).