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time: 10 minutes"><i class="fa fa-clock-o"></i> 10'</span> </ol> <article id="main-article"> <p><em><img alt="" src="../../engineering/torque-1.jpg" style="float: left; width: 250px; height: 152px;">Torque</em>, also known as the <em>moment </em>of a force, is defined as the product of a force and the perpendicular distance from the line of action to the pivot.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <p>Torque is the rotational equivalent of force.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/337018668"></iframe></div> <p>What if the force isn't perpendicular to th<img alt="" src="../../engineering/moment.png" style="background-color: rgb(255, 255, 255); width: 244px; height: 150px; float: right;">e beam?</p> <div class="box"> <p style="text-align: center;"><span class="math-tex">\(\Gamma=Fr\sin \theta\)</span></p> <ul> <li><span class="math-tex">\(\Gamma\)</span> is torque (Nm)</li> <li><span class="math-tex">\(F\)</span> is the force acting (N)</li> <li><span class="math-tex">\(r\)</span> is the distance between the line of action of the force and the pivot</li> <li><span class="math-tex">\(\theta\)</span> is the angle between the force direction and the beam</li> </ul> </div> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/339844045"></iframe></div> </section> <p>A<img alt="" src="../../engineering/couple.png" style="float: left; width: 174px; height: 150px;"> <em>couple</em> is a body with two equal and opposite forces acting on it, at opposite ends. These forces combine to increase the torque on the body.</p> <p>In this case, the total torque acting is:</p> <p><span class="math-tex">\(\Gamma=Fr\sin 90+Fr\sin 90\)</span></p> <p><span class="math-tex">\(\Gamma=2Fr\)</span></p> <hr class="hidden-separator"> <p><img class="sibico" src="../../../img/sibico/video.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="Video"> Watch the effect of a couple on a rod.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/339843914"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <p><em>Moment of inertia</em> is the rotational equivalent of mass. The moment of inertia of a rigid body is defined as the sum of the products of the mass and the squares of their distance from the pivot of all the particles in the body.</p> <div class="box"> <p style="text-align: center;"><span class="math-tex">\(I=\sum mr^2\)</span></p> <ul> <li><span class="math-tex">\(I\)</span> is moment of intertia (kgm<sup>2</sup>)</li> <li><span class="math-tex">\(m\)</span> is the mass of a discrete section of the body (kg)</li> <li><span class="math-tex">\(r\)</span> is the distance between the centre of mass of the discrete section and the pivot (m)</li> </ul> </div> <p>The moment of inertia of some bodies is easy to calculate. You will have to do this for yourself in the exam.</p> <p>For 3 point masses orbitting a common pivot on massless rods, the moment of inertia is the sum of the moment of inertia of the objects: <span class="math-tex">\(I =M_1{r_1}^2+M_2{r_2}^2+M_3{r_3}^2\)</span></p> <p style="text-align: center;"><img alt="" height="251" src="../../screenshot-2019-04-26-at-08.24.03.png" width="280"></p> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p style="text-align: center;"><img alt="" height="317" class="gifffer" data-gifffer="/media/physics/3body.gif" width="265"></p> </section> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/339843969"></iframe></div> </section> <p>For a loop rotating around its centre, all particles can be assumed to be a constant distance from the pivot and so the total mass is at the radius: <span class="math-tex">\(I =mr^2\)</span></p> <p style="text-align: center;"><img alt="" height="256" src="../../screenshot-2019-04-26-at-07.34.11.png" width="259"></p> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p style="text-align: center;"><img alt="" height="270" class="gifffer" data-gifffer="/media/physics/wheel.gif" width="262"></p> </section> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/339843987"></iframe></div> </section> <p>The moment of inertia of more complex <a href="http://en.wikipedia.org/wiki/List_of_moments_of_inertia" target="_blank">3-dimensional shapes</a> is harder to calculate and involves challenging integration from first principles. You will be given the equation for the moment of inertia in the exam. Note that there may be several different moment of intertia values for the same object, depending on the location of the pivot. </p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use flashcards to practise your recall.</em></p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="Normal" data-topics="843" data-subject-id="6" data-n-flashcards="5" style="text-align:center">Show flashcards</a></div><hr> <p><em>Use quizzes to practise application of theory.</em></p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#98abcb6b"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="98abcb6b"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Torque and moment of inertia <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-247-844" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>The moment of inertia of a disc is equal to <span class="math-tex">\({1\over 2} MR^2\)</span>. Which of the following discs, when rotated about the centre, has the largest moment of inertia?</p><p style="text-align: center;"><img alt="" height="317" src="../../engineering/screenshot-2019-07-20-at-08.46.14.png" width="333"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"><span>C</span></label> </p><p><label class="radio"> <input type="radio"><span>D</span></label> </p><p><label class="radio"> <input type="radio">A</label></p><p><label class="radio"> <input type="radio"><span>B</span></label> </p></div><div class="q-explanation"><p>A: <span class="math-tex">\(I={1\over 2}MR^2\)</span></p><p>B: <span class="math-tex">\(I={1\over 2}M({3R\over 4})^2={9\over 32}MR^2\)</span></p><p>C: <span class="math-tex">\(I={1\over 2}({M\over 2})({3R\over 2})^2={9\over 16}MR^2\)</span></p><p>D: <span class="math-tex">\(I={1\over 2}{M\over 3}({4R\over 3})^2={4\over 54}MR^2\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A ship captain exerts equal and opposite forces of 5 N on the outermost points of her 1.5 m helm (steering wheel). What is the resultant torque of this couple?</p><p style="text-align: center;"><img alt="" src="../../engineering/torque-q2.png" style="width: 300px; height: 305px;"></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"><span>5.6 Nm</span></label> </p><p><label class="radio"> <input type="radio"><span>22.5 Nm</span></label> </p><p><label class="radio"> <input type="radio"><span>15 Nm</span></label> </p><p><label class="radio"> <input class="c" type="radio"><span>7.5 Nm</span></label> </p></div><div class="q-explanation"><p>The resultant torque is the sum of each clockwise moment = <span class="math-tex">\(2 \times 5 \text{ N}\times0.75\text{ m}=7.5\text { Nm}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A ladder length 3 m and mass 45 kg rests against a wall at an angle of 40° to the horizontal. Calculate the clockwise moment of the ladder's weight about the point where ladder touches the ground.</p><p style="text-align: center;"><img alt="" src="../../engineering/torque-q3.jpg" style="width: 250px; height: 273px;"> </p></div><div class="q-answer"><p><label class="radio"> <input type="radio"><span>52 Nm</span></label> </p><p><label class="radio"> <input type="radio"><span>500 Nm</span></label> </p><p><label class="radio"> <input class="c" type="radio"><span>520 Nm</span></label> </p><p><label class="radio"> <input type="radio"><span>675 Nm</span></label> </p></div><div class="q-explanation"><p>The line of action of the weight is vertical. The perpendicular distance of this vertical line to the pivot is <span class="math-tex">\(1.5\cos40\)</span>. The moment <span class="math-tex">\(=45\times 10 \times 1.5\cos40=520 \text{ Nm}\)</span>.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram shows two forces acting on a bar.</p><p><img alt="" height="258" src="../../engineering/torqueq1.png" width="358"></p><p>The sum of the torques about C is</p></div><div class="q-answer"><p><label class="radio"><input type="radio"><span class="math-tex">\(F_1(a+b+c) + F_2c\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\(F_1(a+b) + F_2c\)</span></label></p><p><label class="radio"><input class="c" type="radio"><span class="math-tex">\( -F_1(a+b+c) + F_2c\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\(F_1(a+b) - F_2c\)</span></label></p></div><div class="q-explanation"><p>Torque = force x perpendicular distance to pivot. Notice that <span class="math-tex">\(F_1\)</span> produces a clockwise torque and <span class="math-tex">\(F_2\)</span> produces an anticlockwise torque. Another correct answer would have been:<label class="radio"></label></p><p><span class="math-tex">\(F_1(a+b+c)-F_2c\)</span></p><p>HINT: When looking for the distance between the pivot and <span class="math-tex">\(F_1\)</span>, try blocking out <span class="math-tex">\(F_2\)</span> mentally (or with your finger) to ensure that you use the total distance.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram shows two forces acting on a bar.</p><p><img alt="" height="258" src="../../engineering/torqueq1.png" width="358"></p><p>The sum of the torques about A is</p></div><div class="q-answer"><p><label class="radio"><input type="radio"><span class="math-tex">\(F_1(a+b) + F_2(a+b)\)</span></label></p><p><label class="radio"><input class="c" type="radio"><span class="math-tex">\(F_1a + F_2b\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\(F_1a - F_2b\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\(-F_1a + F_2b\)</span></label></p></div><div class="q-explanation"><p>Torque = force x perpendicular distance to pivot. There is no force acting beyond point B, so we can ignore distance <span class="math-tex">\(c\)</span>. Both torques act in a clockwise direction around A.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A force acts on a bar as shown.</p><p><img alt="" height="191" src="../../engineering/totqueq2.png" width="400"></p><p>The torque about A is</p></div><div class="q-answer"><p><label class="radio"><input type="radio"><span class="math-tex">\(FL\cos a\)</span></label></p><p><label class="radio"><input class="c" type="radio"><span class="math-tex">\(FL\sin a\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\(FL\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\(FL \tan a\)</span></label></p></div><div class="q-explanation"><p>The component of <span class="math-tex">\(F\)</span> perpendicular to the bar is <span class="math-tex">\(F\sin a\)</span>.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram shows two forces acting on a bar.</p><p><img alt="" height="258" src="../../engineering/torqueq1.png" width="358"></p><p>If <span class="math-tex">\(F_1= F_2\)</span> the torques would be balanced about point</p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio">None</label></p><p><label class="radio"><input type="radio">A</label></p><p><label class="radio"><input type="radio">B</label></p><p><label class="radio"><input type="radio">C</label></p></div><div class="q-explanation"><p>A trick question - but for good reason! Translationally, the bar would be in linear equilibrium because of equal and opposite <em>forces</em> (remember Mechanics!). Rotationally, there is no position about which these <em>torques</em> would balance unless we could change one of their directions or magnitudes.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A force <span class="math-tex">\(F\)</span> acts on a rod of length <span class="math-tex">\(L\)</span>.</p><p><img alt="" height="116" src="../../engineering/screenshot-2019-07-21-at-09.33.59.png" width="363"></p><p>The torque about A is</p></div><div class="q-answer"><p><label class="radio"><input type="radio"><span class="math-tex">\({3\over 4} FL\)</span></label></p><p><label class="radio"><input class="c" type="radio"><span class="math-tex">\({3\over 5} FL\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\({4\over 5} FL\)</span></label></p><p><label class="radio"><input type="radio"><span class="math-tex">\({4\over 3} FL\)</span></label></p></div><div class="q-explanation"><p>If <span class="math-tex">\(\theta\)</span> is the angle between the force and rod, then <span class="math-tex">\(\Gamma = FL \sin{\theta}\)</span></p><p><span class="math-tex">\(\sinθ = {3\over 5}\)</span> (it's a <span class="math-tex">\(345\)</span> triangle, so this question can be answered without a calculator)</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Four identical small masses are attached to two massless rods as shown.</p><p><img alt="" height="252" src="../../engineering/screenshot-2019-07-22-at-07.19.48.png" width="240"></p><p>The ratio of moment of inertia about axis f to moment of inertia about axis g is</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over2\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(1\over4\)</span></span></label></p></div><div class="q-explanation"><p>Since the masses are small, we can ignore the moment of inertia of the masses on the axis.</p><p>The masses are the same so <span class="math-tex">\(I\)</span> is proportional to <span class="math-tex">\(r^2\)</span>. About axis f, the distance to the masses is half that about axis g, so the moment of inertia is one quarter.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Four identical small masses are attached to two massless rods as shown.</p><p><img alt="" height="266" src="../../engineering/screenshot-2019-07-22-at-07.38.52.png" width="260"></p><p>If the moment of inertia about axis f is <span class="math-tex">\(2mr^2\)</span>, the moment of inertia about axis h is</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4mr^2\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(8mr^2\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(6mr^2\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(10mr^2\)</span></span></label></p></div><div class="q-explanation"><p>Since the masses are small, we can ignore the moment of inertia of the masses on the axis.</p><p>If <span class="math-tex">\(I_f = 2mr^2\)</span> then the mass of blue dots <span class="math-tex">\(=m\)</span> and side of square <span class="math-tex">\(= r\)</span> <span class="math-tex">\(\Rightarrow I_h = 2mr^2 + 2m(2r)^2\)</span>.</p><p>Alternatively, recall the previous response: <span class="math-tex">\(I_h=I_g+I_f=4I_f+I_f\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> 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