Newest questions tagged torque - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-20T03:07:28Z https://physics.stackexchange.com/feeds/tag?tagnames=torque&sort=newest http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/486481 2 Why do the inner tyres of a vehicle rise when taking a turn at high speeds? Raghav Arora https://physics.stackexchange.com/users/234787 2019-06-17T01:33:57Z 2019-06-18T06:22:40Z <p><img src="https://i.stack.imgur.com/hMUrW.jpg" alt="Bus taking turn at high speed. - Speed movie."></p> <p>When a person cycles at high speed on a turn, his body bends inwards, to raise the horizontal component of the normal reaction to provide the centripetal force for the turn. The more the speed is, more is the centripetal force needed and hence more is the bending. </p> <p>Now if in place of the cycle, we keep a bus, then similarly bus needs more centripetal force and hence the normal component should rise by bending the bus inwards. But if we consider common experience, bus bends outwards, raising the inner tyres. Does this not lead to normal reaction being applied away from the centre of the circle? Why does this happen?</p> https://physics.stackexchange.com/q/486460 0 Moment about axis with force applied at non-parallel line user672621 https://physics.stackexchange.com/users/234782 2019-06-16T22:31:09Z 2019-06-16T22:31:09Z <p>I understand the process to solve the moment of a force at a given point applied about an axis. However, what if the force was instead applied to a second, non-parallel?</p> <p>Perhaps a physical example is the moment about the door hinge, if force is evenly applied to a long handle that is contiguously attached to the door, instead of applying it at a door knob? How is the moment about the hinge calculated?</p> https://physics.stackexchange.com/q/485781 0 Can we add torques on different parts of body about a point and calculate total torque about that point? Aramaan meher https://physics.stackexchange.com/users/194928 2019-06-13T03:26:25Z 2019-06-13T03:26:25Z <p>Give proof also . I was doubtful as in a question . Torque due to a couple was calculated using adding torque of the couple due to different parts of the body .</p> https://physics.stackexchange.com/q/485267 1 How to consider a moment of a force as a 2-form? JinsongYang https://physics.stackexchange.com/users/234033 2019-06-10T09:18:08Z 2019-06-12T05:50:34Z <p>My major is mechanical engineering. Recently, I'm reading "The Geometry of Physics An Introduction (3ed)" by Theodore Frankel.</p> <p>On page lix in the section O.r, the author discussed the concept of moment physically and mathematically as quoted below.</p> <blockquote> <p>The moment about the origin, of a force <span class="math-container">$\pmb f$</span> at position vector <span class="math-container">$\pmb r$</span> is, <span class="math-container">$\pmb r \times \pmb f(\pmb r)$</span>, but this expression makes no sense in more than 3 dimensions. But moments and torques surely make sense in any Euclidean <span class="math-container">$\Bbb R^n$</span>, indicating that we have not understood mathematically the notion of moment. Now in cartesian coordinates in <span class="math-container">$\Bbb R^n$</span>, if we replace <span class="math-container">$\pmb r$</span> and <span class="math-container">$\pmb f(\pmb r)$</span> by 1-forms <span class="math-container">$\mathit r=x^adx^a$</span> and <span class="math-container">$\mathit f=f_c(\pmb r)dx^c$</span>, then <span class="math-container">$\mathit r \land f$</span> does make sense as a 2-form at the origin of <span class="math-container">$\Bbb R^n$</span> and its components, in the case of 3 dimensions, coincide with those of <span class="math-container">$\pmb r \times \pmb f(\pmb r)$</span>.</p> </blockquote> <p>Here are my questions:</p> <ol> <li>How to consider the position vector <span class="math-container">$\pmb r$</span> as a 1-form? </li> </ol> <p>EDIT: I can understand that a force can be viewed as a 1-form since its action on a virtual translation (a tangent vector belongs to <span class="math-container">$T_{\pmb r}\Bbb R^3$</span>) gives some virtual work. However, if we consider <span class="math-container">$\pmb r$</span> as a 1-form, I can't figure out on which tangent vector the action of <span class="math-container">$\pmb r$</span> gives a real number. </p> <ol start="2"> <li>If we consider a moment of a force as a 2-form, then how to calculate the virtual work by means of the action of a 2-form on some pair of tangent vectors?</li> </ol> <p>EDIT: In the case of <span class="math-container">$\Bbb R^3$</span>, the moment of <span class="math-container">$\pmb f$</span> is given by <span class="math-container">$\pmb M =\pmb r \times \pmb f(\pmb r)$</span>. Let consider the virtual work done by this moment over a virtual rotation <span class="math-container">$\delta \pmb \theta$</span> (a tangent vector in a tangent space of SO(3)). From the notation of <span class="math-container">$\pmb M$</span>, the virtual work can be viewed as the action of 1-form <span class="math-container">$\pmb M$</span> (a covector in a cotangent space of SO(3)) on the tangent vector <span class="math-container">$\delta \pmb \theta$</span>. When starting from the notation of <span class="math-container">$\pmb r \times \pmb f(\pmb r)$</span>, if we consider the moment as a 2-form, I want to know on which pair of tangent vectors the action of this 2-form gives the same virtual work.</p> https://physics.stackexchange.com/q/485253 2 Conceptual question about wheels Josie Peanut Yael https://physics.stackexchange.com/users/234219 2019-06-10T07:14:41Z 2019-06-12T17:49:25Z <p>I am trying to understand quite naively wheels in a specific framework. </p> <p><strong>Intro</strong> </p> <p>We normally think of squares and circles as different concepts of shapes, but I am reframing it that both are polygons with equal sides. The difference being that a circle is a polygon with infinite equal sides and a square is a polygon with four equal sides.</p> <p>So imagine you have a horse and cart. The cart and its weight is 100kg. The cart has two wheels. In the first case, the cart has square wheels, in the second case round wheels.</p> <p>Let's say the horse pulls the cart over a distance which would make the square wheel do ten complete turns. Call that distance D.</p> <p><strong>Questions</strong></p> <p><strong>1)</strong> <strong>The force that that horse has to pull with in order to make the cart move at a consistent speed across D with</strong> </p> <p><strong>A) the square wheel = polygon, equal sides, number of sides =4</strong></p> <p><strong>B) the round wheel = = polygon, equal sides, number of sides = infinite</strong></p> <p><strong>What would that formula be?</strong> </p> <p>I'd like to know what the universal formula is, referencing the number of sides of the polygon.</p> <p>Then in the case of the round wheel with its infinite sides, simplification of the formula would give I assume a more well-known formula for wheel motion (I don't know what this is)</p> <p><em>(Intuitively, it's clear that there's a far greater load on the square wheel and that also if we were to plot the graph of the force that the horse needs to pull, it would be a maximum right when there is a transition to the side of the square being flat on the ground and it would be a minimum when the line between the center of the square and the corner of the square is 90 degrees to the ground surface. So I imagine the formula would give a wave, and the more/ less sides of the polygon, the more "intense" the wave.. as the number of sides tends to infinity- ie. the wheel is more circular in shape- the wave would dampen to be an absolute number).<br> Guidance would be greatly appreciated!).</em> </p> <p><strong>2) For the above, what the relation to the required maximal axle load? Conceptually and /or formulaically.</strong></p> <p><strong><em>(I'd be happy to know what the formula is but I'd really like to understand the concept better).</em></strong></p> <p>***Intuitively, it's clear that at any point the surface area of the wheel touching the ground- whether the wheel is a circle or square- is exactly the same.</p> <p>NB I am not a physicist- I do understand mathematical concepts and high school physics. Having said that, I'd really be grateful for any explanation that is built ground up rather than quoting things like torque. </p> <p>Thanks!*** </p> https://physics.stackexchange.com/q/483065 0 Conceptual Question Regarding Torque and Acceleration of a system Kevin https://physics.stackexchange.com/users/231827 2019-05-29T04:11:54Z 2019-05-29T04:28:21Z <p>I have a quick question regarding a system with a pulley, a cylinder on an incline, and a block hanging down vertically from the pulley, all connected with a tension cord. IF the pulley has mass, then we consider the torque about that pulley. Then the acceleration of the system would be the same, correct. For an example, in general if a question asks to find the acceleration of the block and the cylinder on the incline, then I can solve for that easily by writing down both the torque formulas and the force formulas while knowing that the acceleration is the same for both.</p> <p>However, if I have a system with the same cylinder on an incline, and a MASSLESS pulley with no inertia, and a block which hung vertically from the incline, then the acceleration of the system wouldnt be the same, correct? For an example, if a question asks what is the acceleration of both the cylinder and the block, I would have to solve each acceleration seperately because they are NOT the same due to a MASSLESS pulley. Please let me know if I am correct or not. A picture of the system is provided below. </p> <p><a href="https://i.stack.imgur.com/KszBk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KszBk.png" alt="enter image description here"></a></p> https://physics.stackexchange.com/q/480529 0 What is the physics behind a pirouette? Mr Pie https://physics.stackexchange.com/users/169739 2019-05-16T23:37:44Z 2019-05-17T11:50:57Z <p>Around last week, I watched a ballet production at the Melbourne Arts Centre, and boy was I amazed! These people dressed up in costumes were spinning on their toes in all kinds of ways, and I was wondering how they did it; i.e. what were the physics behind it.</p> <p>It's called a pirouette, I think, and I went <a href="https://www.youtube.com/watch?v=LLTM_KP0Dsg" rel="nofollow noreferrer">here</a> to learn more about it. The lady there says that males in particular can do up to ten pirouettes, and as the lady demonstrates, this is done on her toes. She also says that the number of turns one does depends on the <em>skill of the dancer</em>. This is not specific and does not give information on the physics behind the pirouette.</p> <p>I know there must be centripetal force created. She said the arms open up, and as she turns, she closes them again, so this must be a drive for momentum. Her right leg is also turned outwards with her foot to her knee (some position called <em>passe</em>) so this might also cause momentum if she is pushing her knee back (which is what it looks like to me). Otherwise the lifted knee would send her leaning towards where it is, but when she turns, she is perfectly straight.</p> <p>However, bringing the arms together and her foot on her knee concentrates a lot of force on just a small base area to work with; that is, the toes. Also, wouldn't torque be playing a role here?</p> <p>If you get a rag and hold it from both ends, it will be pretty loose; but if you stretch it and try to move it, it will be tight and rigid. Maybe the ballet dancer is doing that to control torque? Pushing downwards and lifting upwards at the same time? Newton's third law?</p> <p>Also, why does she start in <em>fourth position</em> for a pirouette? How does she know how much momentum to apply for one and two pirouettes? Why can males do <em>ten</em>?? That's crazy! Does this mean males are more skilled than females?</p> <p>I think a key thing here is <em>weight distribution</em>. This would make a lot of sense, but I am not too sure.</p> <blockquote> <p><strong><em>What is the physics behind a pirouette?</em></strong></p> </blockquote> <hr> <p><sup>If there are any additional related tags, please let me know.</sup></p> https://physics.stackexchange.com/q/480479 1 Physical intuition behind torque converter ffc https://physics.stackexchange.com/users/23316 2019-05-16T18:24:15Z 2019-05-19T19:11:20Z <p>A <a href="https://en.wikipedia.org/wiki/Torque_converter" rel="nofollow noreferrer">torque converter</a> (also <a href="http://web.mit.edu/2.972/www/reports/torque_converter/torque_converter.htm" rel="nofollow noreferrer">here</a>) is a device used in some cars. It uses several "fans" coupled through a liquid (transmission fluid) in order to perform the function of a clutch, but more importantly it acts as a liquid gear in the sense that it multiplies the torque going from the engine to the wheels. </p> <p>Is there an intuitive way to explain what is happening in the liquid? In particular, is it possible to explain the torque multiplication effect without resorting to numerical analysis?</p> https://physics.stackexchange.com/q/480384 2 Why cars have transmission gears? [closed] veronika https://physics.stackexchange.com/users/110669 2019-05-16T07:31:47Z 2019-05-17T07:00:19Z <p>Cars need transmission to efficiently vary the speed of the wheels and this due to the fact that the internal combustion engines have a very limited torque band.</p> <p>What I don't understand is why engines produce a narrow spectre of torque values?</p> https://physics.stackexchange.com/q/480138 0 Magnitude of moment in 3d space Yolanda Hui https://physics.stackexchange.com/users/95122 2019-05-15T04:10:13Z 2019-05-15T04:10:13Z <p><a href="https://i.stack.imgur.com/RGyXy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RGyXy.png" alt="enter image description here"></a></p> <p>What's the magnitude of moment about O, if the vectors F, r1 and r2 all lie in the z-y plane? I know the magnitude of the moment is the cross product of force and the perpendicular distance from the force. So it should be F x r1, but what do you do about r2?</p> https://physics.stackexchange.com/q/479358 0 How is the rotation of a ferromagnetic bullet affected by a coilgun? Nicola Sap https://physics.stackexchange.com/users/155140 2019-05-11T11:23:03Z 2019-05-11T11:23:03Z <p>I've been asked a question that I'm unable to answer. It's about a coilgun, and even if it could probably be expressed in simpler terms, I'll ask it in its entirety just in case I make mistakes while reducing it to simpler terms.</p> <hr> <p>As far as I know, a coilgun (or "coil gun" or "Gauss rifle") is a device that accelerates a ferromagnetic projectile (which is not a permanent magnet) by activating a sequence of solenoidal electromagnets, in such a way that the active solenoid will attract the rod from the core of the previous (now inactive) solenoid, giving it acceleration.</p> <p><a href="https://i.stack.imgur.com/wNK0b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wNK0b.png" alt="enter image description here"></a></p> <p>The question is: if the projectile has a non-zero angular momentum (in the direction of the solenoid axis) at the start, will it be retained after the passage?</p> <p>I don't know much about how magnetic materials work, specifically how the force of attraction of a ferromagnetic object by a magnetic field is modelled, but I believe the question is not as basic as asking whether the solenoid's field has torque, given that the projectile itself alters the field's boundary conditions.</p> https://physics.stackexchange.com/q/479347 2 Friction in pulleys Zam https://physics.stackexchange.com/users/143993 2019-05-11T10:24:34Z 2019-05-11T14:36:49Z <p>The moment equation of a pulley with a rope applying tension on both sides is as follows:</p> <p><span class="math-container">$$I\alpha = f + T_1R - T_2R$$</span></p> <p>( <span class="math-container">$I$</span> - moment of inertia; <span class="math-container">$\alpha$</span> - angular acceleration; <span class="math-container">$f$</span> - friction between axle and the axle holder; <span class="math-container">$T$</span> - Tension; <span class="math-container">$R$</span> - perpendicular distance from centre of axle to tension)</p> <p>The L.H.S and the first term of R.H.S equates to zero as the pulley is 'massless and frictionless'.</p> <p>But there is no mention about friction between rope and pulley. Why is that ?</p> <p>If we consider a pulley to have friction between itself and the ropes, how would its moment equation be ?</p> https://physics.stackexchange.com/q/478593 0 Isn't inertia experience by person in moving bus same as torque Physics freak https://physics.stackexchange.com/users/194320 2019-05-07T19:16:09Z 2019-06-10T04:59:53Z <p>When a bus moves suddenly, the person standing in it tilts backwards. This concept is explained using inertia(tendency of body to resist change in its state of motion) but when the bus moves suddenly, can't we also say that the torque applied due to frictious force on our feet causes our upper body to move backwards?</p> https://physics.stackexchange.com/q/476692 1 Why does the parallel axis theorem apply in this case? jgorton https://physics.stackexchange.com/users/230490 2019-04-29T04:39:24Z 2019-04-29T04:39:24Z <p><a href="https://i.stack.imgur.com/4KrhJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4KrhJ.jpg" alt="question"></a> <a href="https://i.stack.imgur.com/uE3tX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uE3tX.png" alt="initial condition"></a></p> <p>Next, the wheel is flipped over, so the angular momentum of the wheel is now negative. Obviously, the person must start rotating counterclockwise to conserve angular momentum. Since the person is rigidly attached to the shaft, the center of mass of the wheel would start translating in a circle around the AOR.</p> <p><a href="https://i.stack.imgur.com/1OghA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1OghA.png" alt="final"></a></p> <p>This is where the parallel axis theorem comes in. People I have discussed this with, as well as the answer key, believe that <span class="math-container">$\overrightarrow L_f = \omega_f(I_{s,p}+I_w+md^2)-\omega_sI_w$</span> where m is the mass of the wheel, d is the distance from the COM of the wheel to the AOR, <span class="math-container">$I_{s,p}$</span> is the moment of inertia of the person and chair, <span class="math-container">$\omega_f$</span> is the final speed of the person-chair-wheel system, <span class="math-container">$I_w$</span> is the moment of inertia of the wheel about its axis of rotation.</p> <p><a href="https://i.stack.imgur.com/nIML9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nIML9.jpg" alt="answer"></a></p> <p>Here is my rationale: assuming the mass of the wheel is entirely concentrated in its rim, and the bearing it is running on is frictionless, there is no way a vertical torque could be applied to the spinning wheel and thus there would not be a component of the angular momentum related to the wheel rotating around its own center of mass due to the translation of the shaft/wheel CoM (as would be the case for a rigid body; the only components of angular momentum related to the rotation of the wheel are <span class="math-container">$\omega_sI_w$</span> and <span class="math-container">$\omega_fm_wd^2$</span>, and not <span class="math-container">$\omega_fI_w$</span>. This gives me <span class="math-container">$\overrightarrow L_f = \omega_f(I_{s,p}+md^2)-\omega_sI_w$</span>.</p> <p>My understanding is that the parallel axis theorem is only valid for rigid bodies in which the body is rotating around an axis not at its CoM, which this system clearly is not. </p> <p>What is the correct interpretation?</p> https://physics.stackexchange.com/q/476517 2 Is the parallel axis theorem valid for non-rigid bodies? jgorton https://physics.stackexchange.com/users/230490 2019-04-28T05:16:56Z 2019-04-28T18:52:48Z <p>For some context, consider an idealized situation with a person, rigidly attached to the shaft of a bicycle wheel that can spin. sitting on a chair that can rotate. All bearings are frictionless. Initially, the person and chair are at rest but the wheel is spinning, say counterclockwise at rate <span class="math-container">$\omega_s$</span>, around a vertical axis.</p> <p><a href="https://i.stack.imgur.com/uE3tX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uE3tX.png" alt="initial condition"></a></p> <p>Next, the wheel is flipped over, so the angular momentum of the wheel is now negative. Obviously, the person must start rotating counterclockwise to conserve angular momentum. Since the person is rigidly attached to the shaft, the center of mass of the wheel would start translating in a circle around the AOR.</p> <p><a href="https://i.stack.imgur.com/1OghA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1OghA.png" alt="final"></a></p> <p>This is where the parallel axis theorem comes in. People I have discussed this with believe that <span class="math-container">$\overrightarrow L_f = \omega_f(I_{s,p}+I_w+md^2)-\omega_sI_w$</span> where m is the mass of the wheel, d is the distance from the COM of the wheel to the AOR, <span class="math-container">$I_{s,p}$</span> is the moment of inertia of the person and chair, <span class="math-container">$\omega_f$</span> is the final speed of the person-chair-wheel system, <span class="math-container">$I_w$</span> is the moment of inertia of the wheel about its axis of rotation.</p> <p>Here is my rationale: assuming the mass of the wheel is entirely concentrated in its rim, and the bearing it is running on is frictionless, there is no way a vertical torque could be applied to the spinning wheel and thus there would not be a component of the angular momentum related to the wheel rotating around its own center of mass due to the translation of the shaft/wheel CoM (as would be the case for a rigid body; the only components of angular momentum related to the rotation of the wheel are <span class="math-container">$\omega_sI_w$</span> and <span class="math-container">$\omega_fm_wd^2$</span>, and not <span class="math-container">$\omega_fI_w$</span>. This gives me <span class="math-container">$\overrightarrow L_f = \omega_f(I_{s,p}+md^2)-\omega_sI_w$</span>.</p> <p>My understanding is that the parallel axis theorem is only valid for rigid bodies in which the body is rotating around an axis not at its CoM, which this system clearly is not. </p> <p>What is the correct interpretation?</p> https://physics.stackexchange.com/q/476314 3 Is normal reaction a non-central force? rv7 https://physics.stackexchange.com/users/196513 2019-04-27T04:10:04Z 2019-04-27T13:44:54Z <p>For an object of mass m to topple on a rough inclined plane, we consider the torques due to forces acting on it as shown in this figure:</p> <p><a href="https://i.stack.imgur.com/2udBR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2udBR.png" alt="Is normal reaction a non-central force?"></a></p> <p>In the figure, we can see that forces along y-axis are cancelling each other. Hence, net torque about the bottom-right point of box would be, <span class="math-container">$\tau = Wsin\theta \frac{h_{box}}{2}$</span>, which would be non-zero from the beginning of inclination!</p> <p>But that's not true in practical. So, I thought that "normal reaction" should be a non-central force, which is again not true. </p> <p>Help me to resolve this paradox.</p> https://physics.stackexchange.com/q/475896 0 How to find the acceleration of a spool pulled by a force and its work? davidllerenav https://physics.stackexchange.com/users/221731 2019-04-25T06:44:44Z 2019-04-25T15:31:39Z <p>I need some help with this problem:</p> <blockquote> <p>A spool with thread wound on it, of mass <span class="math-container">$m$</span>, rests on a rough horizontal surface. Its moment of inertia relative to its own axis is equal to <span class="math-container">$I= \gamma mR^2$</span> , where <span class="math-container">$\gamma$</span> is a numerical factor, and is the outside radius of the spool. The radius of the wound thread layer is equal to r. The spool is pulled without sliding by the thread with constant force F directed at an angle \alpha to the horizontal. Find:</p> <ul> <li><p>the projection of the acceleration vector of the spool axis on the x-axis.</p></li> <li><p>b)the work performed by the force during the first <span class="math-container">t</span> seconds after the beginning of motion.</p></li> </ul> </blockquote> <p><a href="https://i.stack.imgur.com/OUf5P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OUf5P.png" alt="enter image description here"></a> I already did the first part as shown in the picture below: <a href="https://i.stack.imgur.com/dT8LD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dT8LD.jpg" alt="enter image description here"></a></p> <p>I ended up with <span class="math-container">a=\frac{F(r-R\cos\alpha)}{Rm(\gamma+1)}</span>, but according to my book the answer is <span class="math-container">a=\frac{F(\cos\alpha-\frac{r}{R})}{m(1+\gamma)}</span>. I don't understand why are the signs different, what am I doing wrong?</p> <p>I don't have too much of a clue for the second part, maybe can you give me a hint? Hope you can help me.</p> https://physics.stackexchange.com/q/475865 0 Off-axis torque William Prince https://physics.stackexchange.com/users/211521 2019-04-25T02:26:22Z 2019-04-26T03:28:44Z <p>If one adds pure torque to a disc rotating about a fixed axis in such a way that the application of the torque is off the axis of rotation with the torque vector parallel to the axis of rotation, does the added torque provide angular acceleration to the disc's rotation about the fixed axis regardless of point of application? By pure torque I mean a couple with which no net force is exert on the disc. For example, if a rotating disc had a magnetic dipole attached to its outer edge with the magnetic moment facing the direction of rotation and then a uniform magnetic field was applied to the system such that the dipole's moment were not aligned with the applied field, at that moment in time a torque would be applied to the disc at the point where the dipole was attached. </p> https://physics.stackexchange.com/q/475745 2 Can a tilted wheel roll straight? Rufus https://physics.stackexchange.com/users/138785 2019-04-24T15:16:15Z 2019-04-27T14:18:20Z <p>The common explanation of why a wheel that is falling to one side will turn towards that side to balance itself is gyroscopic precession, i.e. the torque produced by the <strong>falling</strong> of the wheel plus the torque produced by the wheel rolling creates a torque that turns the wheel.</p> <p>My question is what if the wheel was not "falling" but was instead held tilted at an angle to the ground. Would the wheel still roll straight or would it still turn?</p> <p>Example setup (Caster wheels are free to rotate about the vertical axis)</p> <p><a href="https://i.stack.imgur.com/4ghK6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4ghK6.jpg" alt="Setup"></a></p> https://physics.stackexchange.com/q/474630 0 Torque Equation [closed] B V https://physics.stackexchange.com/users/228286 2019-04-23T22:44:43Z 2019-04-24T04:50:34Z <p>There are two torque equations.</p> <p><a href="https://i.stack.imgur.com/bzG2E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bzG2E.png" alt="enter image description here"></a></p> <p>and</p> <p><a href="https://i.stack.imgur.com/MmYUK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MmYUK.png" alt="enter image description here"></a></p> <p>I'm confused about when to use which one given a certain scenario.</p> https://physics.stackexchange.com/q/474266 0 Torque Required To Make Object Circle B V https://physics.stackexchange.com/users/228286 2019-04-22T04:22:17Z 2019-04-22T04:22:17Z <p><a href="https://i.stack.imgur.com/eFLN5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eFLN5.png" alt="enter image description here"></a></p> <p>Say an object is going a constant velocity in the direction of A-B, with respect to its orientation. I'm wondering how you calculate the torque required to always keep point B at the edge of the circle while the object is going clockwise and turning due to the torque. We can say the object has a mass of M.</p> https://physics.stackexchange.com/q/473904 0 Torque in a rubber band Amanda_C https://physics.stackexchange.com/users/224434 2019-04-20T06:36:02Z 2019-04-21T01:26:18Z <p>So I have a bottle that moves with a propeller powered by a rubber band which when you twist the rubber band around and around, the bottle moves.</p> <p>I was wondering how torque plays in this. Is it that the more I twist up the rubber band, the more torque is produced?</p> <p>Would there be a difference in torque if I use a thin rubber band compared to a thick rubber band?</p> <p>Thanks!</p> https://physics.stackexchange.com/q/473554 1 Work from string wrapped around cylinder [closed] Roshan https://physics.stackexchange.com/users/225234 2019-04-18T14:23:24Z 2019-04-18T17:32:15Z <p>A cylinder, on a level surface angularly accelerates as a string wrapped around is put under tension. Because this string causes the cylinder to roll without slipping, the tension force cannot to any work because it is applied to a point on the surface for a zero distance. How then, can the rotational kinetic energy increase if there are no sources of work.</p> https://physics.stackexchange.com/q/473521 0 Bending moment in a cantilever beam Lucifer https://physics.stackexchange.com/users/222265 2019-04-18T11:02:47Z 2019-04-18T17:05:21Z <p>If I have a cantilever beam of length L fixed at the left end to a wall and I hang a weight W from it's right free end then why should the bending moment at a point x units to right of the wall be W(L-x)?</p> <p>If I understand correctly, the bending moment at a point on the beam should be the total torque of the forces acting on cross surface at that point about an axis passing through the geometric center and perpendicular to the plane of bending, then how is this equal to W(L-x)?</p> https://physics.stackexchange.com/q/473439 1 Extending general relativity with torque based on quasimetrics David Jonsson https://physics.stackexchange.com/users/68701 2019-04-18T00:21:12Z 2019-04-18T00:21:12Z <p>If torque is allowed to exist in the space part of the stress energy tensor <span class="math-container">T_{\mu\nu}</span> in the Einstein field equations <span class="math-container"> R_{\mu\nu}-\frac12 g_{\mu\nu} R = 8\pi T_{\mu\nu} </span> it would lead to <span class="math-container">$T_{\mu\nu}$</span> being asymmetric. That would require some tensor in the left hand side of the field equations to be asymmetric as well. The Ricci tensor <span class="math-container">$R_{\mu\nu}$</span> can hardly be generalized to become asymmetric. However the metric tensor <span class="math-container">$g_{\mu\nu}$</span> can become asymmetric by using a <a href="https://en.wikipedia.org/wiki/Metric_(mathematics)#Quasimetrics" rel="nofollow noreferrer">quasimetric</a> where the symmetry criteria is dropped meaning that the distance from a to b is different from the distance from b to a. This can be further imagined as taking the gauge freedom in the Lorentz transformation to act on the speed of light, as is done in the <a href="https://en.wikipedia.org/wiki/Sagnac_effect" rel="nofollow noreferrer">Sagnac effect</a> where a longer distance is experienced when measuring in the direction of a motion instead of measuring against a motion. The distance is measured symmetrically in the stationary reference frame.</p> <p>Could such an extension of the field equations be made theoretically and tried on a real physical situation?</p> https://physics.stackexchange.com/q/473409 0 Torque require to rotate a shaft Joaquin Osses https://physics.stackexchange.com/users/179244 2019-04-17T21:00:52Z 2019-04-17T21:32:12Z <p><a href="https://i.stack.imgur.com/VuH1b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VuH1b.png" alt="enter image description here"></a></p> <p>I have a solid cylinder (shaft) that I want to rotate around its axis. The cylinder is supported by two bearings at its ends. No other loads than the weight are applied to the cylinder. My question is; how much torque it needed to rotate the shaft? I don't know how to applied the friction coef. of the GT2 and bearing. 2. how to calculate the torque required to maintain the cylinder rotating at the desired speed once it reaches it. From what I understood, it is dependent on the friction at the bearings only.</p> https://physics.stackexchange.com/q/471698 1 Normal reactions with moments ramose https://physics.stackexchange.com/users/224610 2019-04-10T10:12:12Z 2019-04-10T12:14:01Z <blockquote> <p><a href="https://i.stack.imgur.com/HLZLU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HLZLU.png" alt="enter image description here"></a> A uniform ladder, <span class="math-container">$AB$</span>, is leaning against a smooth vertical wall on rough horizontal ground at an angle of <span class="math-container">$70°$</span> to the horizontal. The ladder has length <span class="math-container">$8\ \rm m$</span>, and is held in equilibrium by a frictional force of magnitude <span class="math-container">$60\ \rm N$</span> acting horizontally at <span class="math-container">$B$</span>, as shown in the diagram. Write down the magnitude of the normal.</p> </blockquote> <p>The normal reaction makes both A and B rotate, so I don't understand how we can ignore it by taking the moment at B OR A?</p> <p>I have 60*8cos20 on the LHS (clockwise) = X*9.8*4sin20 (anti) but then I remember the normal force and it feels like I should add everything from the LHS to the RHS and I don't understand why that's wrong.</p> <p>If I'm taking the moment at B, the 60N to the left frictional force still makes point -A rotate clockwise. and the Normal force, makes point A rotate anti-clockwise, so I still can't ignore it?</p> <p>Similarly, there have been other seesaw questions, where I am told if I take the moment at the pivot then I can ignore the reaction at the pivot. Sure, but isn't there still a normal reaction from the seesaw back up into the (particle) weights on top of them?</p> <p>Please can someone explain the simple thing I must be doing incorrectly, thanks!</p> https://physics.stackexchange.com/q/471036 0 Torque from Newton’s Laws Steve https://physics.stackexchange.com/users/197788 2019-04-07T06:33:24Z 2019-04-07T06:33:24Z <p>Is is possible to predict the motion of rigid bodies only in terms of Newton’s laws of motion without torque (for example by using the system of particles model)?</p> <p>For instance, if there was a rod of length ‘l’ attached to a hinge, and a force ‘F’ was applied (perpendicular to the rod) a distance ‘r’ away from the hinge, then can the rod be broken up into infinitesimal masses (particles) then work out the internal and external forces on the rod (forces on each particle) to predict its angular acceleration. Then from this prediction maybe the torque=(moment of inertia)*(angular acceleration) relationship can be derived in terms of Newton’s laws.</p> https://physics.stackexchange.com/q/469822 0 Counter Torque in Electric Generator Sanat Kumar https://physics.stackexchange.com/users/226977 2019-04-01T06:00:37Z 2019-04-01T06:00:37Z <p>I am going to use some sort of “dynamo” or electric generator component to generate electricity from the flowing drilling fluid. However, the effect of the magnet rotating within an electrical coil creates a “torque” or resistance to the motion of the magnet when current is drawn from the coil. This resistance or counter torque will affect the drilling, so I need to calculate the maximum counter torque that we would generate in this system. The information available to me at the moment is that I would need about 1000 W of power and the generation voltage at the generator is roughly 20-120 V. You may assume the drill bit (or mud motor and magnet in the electrical generator) is rotating at 60 Hz. For all other parameters, you may pick reasonable values.</p> https://physics.stackexchange.com/q/469602 1 Angular Impulse equations for rotational dynamics programmer https://physics.stackexchange.com/users/226550 2019-03-30T20:02:04Z 2019-03-31T04:48:09Z <p>Consider a 3d rigid body at rest initially, assuming no net external forces acting on it. It is set in perpetual rotation along one of its principal axis. Now an angular impulse (which is a vector) acts on it in certain direction not parallel to its principal axis (the one parallel to axis of rotation).</p> <p>we have two observations:</p> <ol> <li>from newtons second law of motion analogous on rotational dynamics, we have <span class="math-container">$\mathbf \tau = \frac{\text d\mathbf L}{\text dt}$</span>, or <span class="math-container">$\int \mathbf\tau\ \text dt = \Delta\mathbf L$</span> (<span class="math-container">$\mathbf\tau$</span>, <span class="math-container">$\mathbf L$</span> are torque and angular momentum vectors) </li> <li>finally after the angular impulse acts on it the body still remains in perpetual rotation about some principal axes.</li> </ol> <p>How would you find the new axis of rotation ?</p>