Question

A flywheel in the form of a uniformly thick disk of radius 1.88 m, has a mass of 36.6 kg and spins counterclockwise at 371 rpm. Calculate the constant torque required to stop it in 2.75 min.

Answer 1

Concepts and reason

The required concepts to solve these questions are rotational kinematic equations, moment of inertia and torque.

Firstly, calculate the moment of inertia, calculate the angular acceleration by using rotational kinematic equations and then calculate the required torque to stop the car.

Fundamentals

The expression for angular velocity is expresses as follows,

$\omega = {\omega _0} + \alpha t$

Here, $\alpha$is the angular acceleration, $t$ is the time and ${\omega _0}$is the initial angular velocity.

The expression for moment of inertia for disk is expresses as follows,

$I = \frac{1}{2}m{R^2}$

Here, $R$is radius of the disk and $m$is the mass of the flywheel.

The expression for torque is as follows,

$\tau = I\alpha$

Here, $\alpha$is the angular acceleration and $I$is the moment of inertia.

The expression for moment of inertia is,

$I = \frac{1}{2}m{R^2}$

Substitute $1.88{\rm{ m}}$for $R$and $36.6{\rm{ kg}}$for $m$in the above equation.

$\begin{array}{c}\\I = \frac{1}{2}\left( {36.6{\rm{ kg}}} \right){\left( {1.88{\rm{ m}}} \right)^2}\\\\ = 64.67{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The expression for angular velocity is as follows,

$\omega = {\omega _0} + \alpha t$ …… (1)

Rearrange the equation for angular acceleration.

$\alpha = \frac{{\omega - {\omega _0}}}{t}$ …… (2)

Convert the unit of $\left( {{\omega _0}} \right)$initial angular velocity $\left( {371{\rm{ rpm}}} \right)$to${\rm{rad}} \cdot {{\rm{s}}^{ - 1}}$.

$\begin{array}{c}\\371{\rm{ rpm}} = \left( {371{\rm{ rpm}}} \right)\left( {\frac{{0.104{\rm{ rad}} \cdot {{\rm{s}}^{ - 1}}}}{{1{\rm{ rpm}}}}} \right)\\\\ = 38.58{\rm{ rad}} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

Substitute $0$ for $\omega$ , $38.58{\rm{ rad}} \cdot {{\rm{s}}^{ - 1}}$ for ${\omega _0}$ and$2.75{\rm{ min}}$ for $t$in the equation (2).

$\begin{array}{c}\\\alpha = \frac{{0 - \left( {38.58{\rm{ rad}} \cdot {{\rm{s}}^{ - 1}}} \right)}}{{\left( {2.75{\rm{ min}}} \right)\left( {\frac{{60{\rm{ s}}}}{{1{\rm{ min}}}}} \right)}}\\\\ = - 0.23{\rm{ rad}} \cdot {{\rm{s}}^{ - 2}}\\\end{array}$

The expression for torque is as follows,

$\tau = I\alpha$

Substitute $64.67{\rm{ kg}} \cdot {{\rm{m}}^2}$ for $I$ , $0.23{\rm{ rad}} \cdot {{\rm{s}}^{ - 2}}$ for $\alpha$ in the above equation of torque.

$\begin{array}{c}\\\tau = \left( {64.67{\rm{ kg}} \cdot {{\rm{m}}^2}} \right)\left( { - 0.23{\rm{ rad}} \cdot {{\rm{s}}^{ - 2}}} \right)\\\\ = - 14.87{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

The magnitude of the torque is,

$\tau = 14.87{\rm{ N}} \cdot {\rm{m}}$

Ans:

The required torque to stop the wheel is $14.87{\rm{ N}} \cdot {\rm{m}}$.