Question

5. The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00s, starting from rest? ANS: 13.1 N m

Answer 1

Concept and reason

The concept needed to solve this problem is torque.

Initially, write an expression for angular acceleration in terms of initial angular speed, final angular speed, and time. Calculate angular deceleration of the fly wheel. Later, write the equation for torque in terms of moment of inertia and angular acceleration. Finally, calculate the torque needed to stop the flywheel.

Fundamentals

The expression for angular acceleration ( $\alpha$ ) in terms of initial angular speed $\left( {{\omega _0}} \right)$ , final angular speed $\left( \omega \right)$ , and time interval t is,

$\alpha = \frac{{\omega - {\omega _0}}}{t}$

The expression for the torque in terms of moment of inertia (I) and angular acceleration is,

$\tau = I\alpha$

The expression for angular acceleration is,

$\alpha = \frac{{\omega - {\omega _0}}}{t}$

Convert the unit of the final angular speed of the flywheel from rev/min to rad/s.

$\begin{array}{c}\\\omega = 400{\rm{ rev/min}}\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right)\left( {\frac{{1{\rm{ min}}}}{{60{\rm{ s}}}}} \right)\\\\ = {\rm{41}}{\rm{.87 rad/s}}\\\end{array}$

Substitute 0 rad/s for ${\omega _0}$ , ${\rm{41}}{\rm{.87 rad/s}}$ for $\omega$ , and 8.00 s for t in the equation.

$\begin{array}{c}\\\alpha = \frac{{{\rm{41}}{\rm{.87 rad/s}} - 0{\rm{ rad/s}}}}{{8.00{\rm{ s}}}}\\\\ = {\rm{5}}{\rm{.23 rad/}}{{\rm{s}}^2}\\\end{array}$

The expression for the torque in terms of moment of inertia and angular acceleration is,

$\tau = I\alpha$

Substitute $2.50{\rm{ kg}} \cdot {{\rm{m}}^2}$ for I and ${\rm{5}}{\rm{.23 rad/}}{{\rm{s}}^2}$ for $\alpha$ .

$\begin{array}{c}\\\tau = \left( {2.50{\rm{ kg}} \cdot {{\rm{m}}^2}} \right)\left( {{\rm{5}}{\rm{.23 rad/}}{{\rm{s}}^2}} \right)\\\\ = {\rm{13}}{\rm{.08 N}} \cdot {\rm{m}}\\\\ = 13.1{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans:

The required value of torque is $13.1{\rm{ N}} \cdot {\rm{m}}$ .

Answer 2

first we have to find out the angular acceleration in 8sec using rotational equation of motion. Then simply put in the formula of torque .

Answer! angulare final accelereation intal OM Woa angulare x keed=0 You een min Wo Yoox gt rad = 41.89 reads 60 الم fa 88 wa wotat = 0+0 (8) = 5.2 ead/22 sog 41.89 Thon, torque to IX = 2.50x 5.24 = - 13.1Nm