Question

A pulley having a rotational inertia of \(1.14 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(9.88 \mathrm{~cm}\) is acted on by a force, applied tangentially at its rim, that varies in time as \(F=A t+B t^{2}\), where \(A=0.496 \mathrm{~N} / \mathrm{s}\) and \(B=0.305 \mathrm{~N} / \mathrm{s}^{2} .\) If the pulley was initially at rest, find its angular speed after \(3.60 \mathrm{~s}\).

Short answer

The final answer will be in rad/s after calculating the definite integral in the last step.

Step-by-step solution

Calculation of Net Torque

The torque on the pulley can be calculated by multiplying the force by the radius. However, given that the force varies with time - \(F = At + Bt^2\), we need to express the torque as a function of time as well. We convert the radius from centimeter to meter for the correct SI unit. So, the net torque, \(\tau\), is: \(\tau = rF = r(At + Bt^2)\), where \(r = 9.88/100 = 0.0988m\).

Calculation of Angular Acceleration

The angular acceleration, represented by \(\alpha\), can be calculated by dividing the net torque by the rotational inertia. So, \(\alpha = \frac{(At + Bt^2)r}{I}\), where \(I = 1.14 \times 10^{-3} kgm^2\).

Calculate Angular Velocity

The angular speed, represented by \(\omega\), can be calculated by multiplying the angular acceleration by the time, given that the initial angular speed is zero. Here, the time is given as \(t = 3.60s\). So, \(\omega = \int_0^3.60 \alpha dt = \int_0^3.60 \frac{(At + Bt^2)r}{I} dt \).

Key concepts

Understanding Torque

In rotational dynamics, torque is a fundamental concept akin to force in linear dynamics. It tells us how effectively a force can cause an object to rotate. Torque is not just about applying a force; the position where the force is applied and the angle relative to the pivot point are crucial. It is calculated by the formula \( \tau = rF \cdot \sin(\theta) \), where \(r\) is the distance from the pivot point (radius in the case of a pulley), \(F\) is the magnitude of the force, and \(\theta\) is the angle between the force and the lever arm.

In the exercise, the torque is simplified because the force is perpendicular to the radius (\( \theta = 90^\circ \), making \( \sin(\theta) = 1 \)). This simplifies the torque calculation to \( \tau = rF \). The force \( F \) isn't constant but depends on time as given by \( F = At + Bt^2 \). Therefore, the torque changes over time, written as \( \tau(t) = r(At + Bt^2) \). This underlines the nature of rotational motion being dynamic and often non-uniform.

Exploring Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time, similar to linear acceleration but in rotational terms. It is symbolized by \( \alpha \), and its units are typically radians per second squared (rad/s²). Angular acceleration can be determined from the net torque and the moment of inertia by using the formula \( \alpha = \frac{\tau}{I} \), where \( I \) is the rotational inertia, representing how much torque is needed for a certain angular acceleration.

In the problem, the rotational inertia \( I \) is given as a constant, \( 1.14 \times 10^{-3} \text{ kg m}^2 \), and the torque varies over time, expressed as \( \tau = r(At + Bt^2) \). So, the angular acceleration also varies over time: \( \alpha(t) = \frac{r(At + Bt^2)}{I} \). This time-dependent aspect makes problems in rotational dynamics more complex, as it requires integrating over time to find other quantities like angular velocity.

Angular Velocity Insight

Angular velocity represents how fast an object rotates and is measured in radians per second (rad/s). It's the rotational counterpart to linear velocity, showing not just speed but direction around a circular path. The initial conditions of a rotational system often greatly impact the resulting angular motion. In this exercise, we are given that the pulley starts from rest, meaning the initial angular velocity is zero.

We calculate the angular velocity \( \omega \) by integrating the angular acceleration \( \alpha \) over time: \( \omega = \int_0^{t} \alpha \, dt \). Due to the time-dependent nature of \( \alpha = \frac{r(At + Bt^2)}{I} \), this involves setting up an integral that accounts for the changes in force as time progresses. Evaluating this integral from \( t = 0 \) to \( t = 3.60 \text{s} \) allows us to find the angular speed at the end of the specified time period, giving insight into how such systems gain speed through rotational energy inputs.