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Perpendicular Force = No Speed Change

  • Thread starter Swapnil
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Suppose an object is moving at some veocity v. Why is it that if you apply a force in the direction perpendicular to the direction of velocity, you only change the direction of the velocity and not its magnitude.

I am having a feeling that I am gonna get one of those "that's just the way it is" answers to this question...:rolleyes:
 

Chi Meson

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If you pull on something such that the direction of pull is in the same direction that it is moving, then you'd expect the object to go faster, right?

And if you pull on something in the opposite direction, then the object would slow down.

If you pull on something so that there is a componant of your force along the direction of motion, and another componant perpendiculat to the direction of motion, then the object would speed up slightly, and also change direction slightly (as it accelerated in the direction of the perpendicular componant).

If your pull has a componant that is opposite motion, plus a componant that is perpendicular to motion, then it would slow down slightly, and also chage direction.

SO, if there is no componant either in the direction of motion, nor opposite, then the object won't speed up or slow down. In the case of circular motion, every moment the direction of force changes slightly so that the force is always perpendicular to the moment of motion.
 
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You have given me an intuitive answer; I already know why it makes sense intuitively. But the is there is a more mathematical answer to my question?
 

HallsofIvy

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"speed" is the length of the velocity vector [itex]s= \sqrt{\vec{v}\cdot\vec{v}}[/itex]. Differentiating with respect to time,
[tex]\frac{ds}{dt}= \frac{1}{\sqrt{\vec{v}\cdot\vec{v}}}\vec{v}\cdot\frac{d\vec{v}}{dt}[/tex]
by the product rule.
Of course, [itex]\frac{d\vec{v}}{dt}[/itex] is the acceleration. If the force is perpendicular to [itex]\vec{v}[/itex], then so is the acceleration (acceleration= Force/mass) so that product [itex]\vec{v}\cdot\frac{d\vec{v}}{dt}[/itex] is 0. Then entire derivative then is 0: [itex]\frac{ds}{dt}= 0[/itex].

A somewhat more physical explanation is that if there is no component of force in the direction of motion, then there is no work done so the kinetic energy cannot change.
 

arildno

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Remember that kinetic energy is a scalar, rather than a vector. Thus, forces that only change the object's velocity, but not its speed, cannot influence the object's kinetic energy.
 
459
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HallsofIvy said:
"speed" is the length of the velocity vector [itex]s= \sqrt{\vec{v}\cdot\vec{v}}[/itex]. Differentiating with respect to time,
[tex]\frac{ds}{dt}= \frac{1}{\sqrt{\vec{v}\cdot\vec{v}}}\vec{v}\cdot\frac{d\vec{v}}{dt}[/tex]
by the product rule.
Of course, [itex]\frac{d\vec{v}}{dt}[/itex] is the acceleration. If the force is perpendicular to [itex]\vec{v}[/itex], then so is the acceleration (acceleration= Force/mass) so that product [itex]\vec{v}\cdot\frac{d\vec{v}}{dt}[/itex] is 0. Then entire derivative then is 0: [itex]\frac{ds}{dt}= 0[/itex].
What about the fact that the force perpendicular to the direction of the velocity changes the direction of the velocity?
 

Doc Al

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Swapnil said:
What about the fact that the force perpendicular to the direction of the velocity changes the direction of the velocity?
In general, acceleration ([itex]d\vec{v}/dt[/itex]) involves changes in both speed and direction. But as Hall's showed, when the acceleration is perpendicular to velocity the change in speed is zero. All that's left is a change in direction.
 

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