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## New answers tagged torque

0

I magnetic dipole can be modelled by a small coil of wire. The dipole moment of such a coil is defined to be $$\vec m=I\vec A$$ Where $I$ is the current and $\vec A$ is the area vector (that is the area multiplied by a unit vector perpendicular to the plane. The torque on a dipole is the sum of $$\vec \tau =\vec r \times \vec F$$ for each force $\vec F$ ...

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I want to give an answer for the case, that the mentioned particles are electrons. Let us consider that the magnetic dipole moments of this two electrons are aligned in a straight line through the points (0,a,0) and (a,a,0). Since both electrons are moving their magnetic dipole moments begin to turn when the electrons leave the mentioned points. Perhaps it ...

1

EDIT: The "old" explanation below is quite informal, as gravity cannot produce a torque around the center of mass. See the new explanation below it. Humans lean forwards slightly as they run (more so if they run faster, and have to counter larger drag forces). This makes gravity exert a counter-torque that exactly balances the torque produced by action of ...

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1) first thing you have to notice is wind loads on panels is far more than dead weight of panels 2) so the torque required mostly depends on peak wind load. 3)It also depends on friction in bearing. 4)Finally the weight of the panels should be considered, lower the moment of inertia about rotational axis, lower the torque required.

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The cited diagram depicts but a small part of the motions that occur during a dive. That appears to be an overly simplified diagram of a rather simple springboard dive, a forward 1 1/2 somersault tuck. It's overly simplified because it misses the complex motions a diver undergoes at the start of the dive. There are two mechanisms by which a diver turns ...

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The equation from the link is an approximation. In your analysis you do not include the mass of the car when considering linear acceleration. Nor do you consider the reaction force of the wheel/axle accelerating the car. You can split the driving torque into two portions, the portion that rotationally accelerates the wheel, and the portion that is ballanced ...

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The force doesn't get "used up" by creating a torque. The torque and the force exist simultaneously. You correctly computed the torque due to that force. If that is the only force on this object, then the linear acceleration of the object is given by Newton's 2nd law: $$a = \frac{100~\rm N}{m}$$ where $m$ is the mass of the object. That is the linear ...

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If you have a single particle then you can indeed easily describe its motion using Newton's laws. However in rotating systems we are often dealing with continuous bodies not point particles. These are characterised by a moment of inertia, rather than just by a mass, and we have an equation analogous to Newton's second law: $$T = I\dot{\omega} = ... 0 This problem or rather question is very similar to a problem I solved in actuating a mirror's angle in two degrees of freedom (yaw and pitch) using a central fulcrum or pivot by which the mirror rotated, and three piezoelectric actuators that acted 120 degrees apart at the mirror's circular periphery. To transform three axis actuation to two axis orthogonal ... 1 You need to know where the center of mass is for the three masses (on the three arms). This is of course just the position weighted mean - so if the scales are at position \vec r_i, and contain mass m_i, then the center of mass is at$$\vec C = \frac{\sum{m_i \vec r_i}}{\sum{m_i}}$$Note that we are using vector addition here - this is really just a 2D ... 1 So the torque from the back wheel causes a net torque about the centre of mass counterclockwise. This means the back wheel exerts a greater reaction force on the ground so when we resolve forces the reaction force exceeds mg, causing a net force up on the centre of mass. 0 Sorry if I'm missing a detail here - but how can you choose the center of mass? Isn't that determined by the shape/mass distribution of the bike? And the axis of rotation ought to be the center of the rear tire, no? Just imagining a wheelie I can see that the center of mass should move upwards to some degree, as the wheelie is an upward rotation. (posting ... 1 You want to consider the following factors: Maximize initial energy stored in the rubber band. This means you need to be able to twist the band lots of times, and as it unwinds it must continue to produce torque. Minimize internal friction - make the mechanism that converts power from the band to kinetic energy as "direct" as possible. In fact a rubber ... 1 You are really asking about the reaction forces felt on a rod when you push on its end. For a rod, you can work with a quantity called the reduced mass (see for example this excellent answer for the derivation). As long as the rod is balanced on its end, the reduced mass tells you exactly how much greater the inertia of the cart appears to be:$$m_r = ...

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Consider torque as a force applied to a lever connected to a rotation point. Torque from the piston to the rotating center of the crankshaft is internal to the engine and will not affect the motorcycle unless the engine is connected by a power train to another rotation point on the motorcycle - the rear axle. The motorcycle will NOT try to rotate about its ...

4

If we take the free body diagram above and sum the moments about the center of mass, we would find that an increased applied force would in fact cause the solid body to rotate. Perhaps. Or additional forces can appear. If I push up on my car's bumper, a rotational force is being applied. But the normal force on the wheel farther from me ...

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No. You have to take moments about the center. Taking moments clockwise, 252Nm + 240Nm - 246Nm - 198Nm = 48Nm Therefore it has a clockwise moment of 48Nm which implies it moves in the clockwise direction.

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