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- #2

Chestermiller

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It would change the direction of the velocity vector, but not its magnitude.

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It provides a little change of velocity, and if the force keeps acting on the object it may do the constant-speed circular motion, for it doesn't give the object acceleration parallel to itself.Wouldn't the force give rise to a vertical component of velocity on the object and thus increase the speed of the object?

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Khashishi

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A.T.

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How should the speed change, increase or decrease?If a force acts on an object perpendicular to its velocity, why does the speed remains unchanged?

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To prove this, take P to be the power of the force delivered when the object has velocity V,

Then P= (Force vector).(Velocity vector) at that instant, but since Force and velocity are perpendicular, P would be 0.

Hence work done by the force would be 0, and hence change in Kinetic energy would be zero.Thus, the speed would remain same, but the direction of velocity changes.

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- #8

A.T.

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Because you cannot apply a force perpendicularly to a velocity of zero.Let's suppose the object is stationary. If the force acts on it in any direction it will give rise to a SPEED in that direction. Why cant we apply the same analogy in circular motion?

Speed doesn’t have components, it's a scalar.More importantly, why isn't force giving rise to a component of SPEED in any direction?

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sophiecentaur

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Energy considerations are often a good way into solving problems and explaining things.

Here's something to think about. You are thinking that, perhaps, speed should increase. If the speed increased then so would the Kinetic Energy. That would involve the need for Energy to be put into the system. (in the absence of friction, circular motion will carry on forever) The only way that energy to be put in would be in the form of Work, which is Force times the distance moved along the line of action of the force. The Force would be the centripetal force and the distance (when moving in a circle) would be Zero (radius doesn't change). So no work is done and, hence, the Kinetic Energy (and consequently the speed) can't be increasing.

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jbriggs444

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You may be imagining a long, narrow right triangle with the velocity of the object as the long side, the incremental velocity produced by acceleration as the short side and the resulting final velocity as the hypotenuse. That picture is problematic for three reasons (or one reason expressed in three ways perhaps).

1. In the limit as the size of the interval goes to zero, the hypotenuse is the same length as the initial velocity (mentioned by Khashishi)

2. The acceleration is in a constantly changing direction. It's not really a right triangle because the current acceleration not always at right angles to the initial velocity. The short side is a circular arc whose direction is ambiguous.

3. The picture is prejudiced toward progress forward in time. If you extend it backwards to a tiny prior interval, the initial velocity is the hypotenuse, the increment is the short side and the final velocity is the long side. Rather than speeding it up, the mental picture says that the acceleration should be slowing the object down.

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Chestermiller

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The force will be drawing the object towards the center. So as soon the object will change direction and try to orient towards the center it would've travelled a small distance in the direction of force.Energy considerations are often a good way into solving problems and explaining things.

Here's something to think about. You are thinking that, perhaps, speed should increase. If the speed increased then so would the Kinetic Energy. That would involve the need for Energy to be put into the system. (in the absence of friction, circular motion will carry on forever) The only way that energy to be put in would be in the form of Work, which is Force times the distance moved along the line of action of the force. The Force would be the centripetal force and the distance (when moving in a circle) would be Zero (radius doesn't change). So no work is done and, hence, the Kinetic Energy (and consequently the speed) can't be increasing.

- #13

Chestermiller

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So???The force will be drawing the object towards the center. So as soon the object will change direction and try to orient towards the center it would've travelled a small distance in the direction of force.

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I am not talking about the triangle because when the angle is very small it can resemble a right triangle. What I am asking is whenever we apply a force the objects is accelerated in that direction. As a result the magnitude of velocity must change. And if it isn't changing then it must be because the other component of the velocity is decreased. So what is decreasing the other component?You may be imagining a long, narrow right triangle with the velocity of the object as the long side, the incremental velocity produced by acceleration as the short side and the resulting final velocity as the hypotenuse. That picture is problematic for three reasons (or one reason expressed in three ways perhaps).

1. In the limit as the size of the interval goes to zero, the hypotenuse is the same length as the initial velocity (mentioned by Khashishi)

2. The acceleration is in a constantly changing direction. It's not really a right triangle because the current acceleration not always at right angles to the initial velocity. The short side is a circular arc whose direction is ambiguous.

3. The picture is prejudiced toward progress forward in time. If you extend it backwards to a tiny prior interval, the initial velocity is the hypotenuse, the increment is the short side and the final velocity is the long side. Rather than speeding it up, the mental picture says that the acceleration should be slowing the object down.

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There will be a gain in energy which could compensate for the increase in kinetic energy.So???

- #16

sophiecentaur

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Yes it is but the radius does not decrease (it doesn't get any closer to the centre) so no Work is done. If you were sitting at the centre of rotation and going round at the same rate, the string you were holding would not be changing length and the tension in the string would be the same. No change in energy so no change in speed,The force will be drawing the object towards the center. So as soon the object will change direction and try to orient towards the center it would've travelled a small distance in the direction of force.

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sophiecentaur

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You are basically asking how the Calculus method can be used to establish the vI am not talking about the triangle because when the angle is very small it can resemble a right triangle. What I am asking is whenever we apply a force the objects is accelerated in that direction. As a result the magnitude of velocity must change. And if it isn't changing then it must be because the other component of the velocity is decreased. So what is decreasing the other component?

- #18

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My question is simply why is the magnitude of velocity not changing even though a force is acting on the body?

I am going to list why I am confused about this.

1. If a force acts on an object, the object is accelerated. What I think it means is it gets velocity in that direction and the other components of velocities on that object are not influenced by that force. (Components of velocity perpendicular to the force)

2. Since an object is accelerated and gains a "new" velocity in that direction, the resultant velocity will be bound to be greater

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sophiecentaur

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How can there be any change in the Energy if the motion is not speeding up or slowing down?there could be an increase in energy.

[or vice versa].

- #20

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I can comprehend calculus but calculus wasn't used to derive the formula vYou are basically asking how the Calculus method can be used to establish the v^{2}/r formula. If you understand any calculus then the answer is easy. If not then it may be best to accept it and find out later how to derive the formula. Physics will be full of formulae that you cannot yet derive but you are happy to accept the results - so don't feel bad about it. An arm waving explanation is not really a 'proof' in any case.

This was how it was derived

dx = dv/v (assuming that angle x is very small )

vdx = dv

v dx/dt = dv/dt

Then by substituting the necessary equation we arrived at a = v

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jbriggs444

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No. It does not. Not at all. It travelsBut the object does travel in the direction of forc

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sophiecentaur

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How can it, if the string doesn't get shorter?But the object does travel in the direction of force

At this time, I think you need to find reasons why you may be wrong, rather than in still wanting to be right. There is something in your head that is not letting you accept the 'correct' answer.

- #23

sophiecentaur

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How can you say that calculus was not used? Your last post was full of it.I can comprehend calculus but calculus wasn't used to derive the formula v^{2}/r (when I was taught)

This was how it was derived

dx = dv/v (assuming that angle x is very small )

vdx = dv

v dx/dt = dv/dt

Then by substituting the necessary equation we arrived at a = v^{2}/r

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- #25

jbriggs444

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A force with a constant direction is not centripetal.

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