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Faiq
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If a force acts on an object perpendicular to its velocity, why does the speed remains unchanged? Wouldn't the force give rise to a vertical component of velocity on the object and thus increase the speed of the object?
It would change the direction of the velocity vector, but not its magnitude.Faiq said:If a force acts on an object perpendicular to its velocity, why does the speed remains unchanged? Wouldn't the force give rise to a vertical component of velocity on the object and thus increase the speed of the object?
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.Faiq said:Wouldn't the force give rise to a vertical component of velocity on the object and thus increase the speed of the object?
How should the speed change, increase or decrease?Faiq said:If a force acts on an object perpendicular to its velocity, why does the speed remains unchanged?
Because you cannot apply a force perpendicularly to a velocity of zero.Faiq said: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 can't we apply the same analogy in circular motion?
Speed doesn’t have components, it's a scalar.Faiq said:More importantly, why isn't force giving rise to a component of SPEED in any direction?
Energy considerations are often a good way into solving problems and explaining things.Faiq said:I know the velocity is changing because of the direction. 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 can't we apply the same analogy in circular motion? More importantly, why isn't force giving rise to a component of SPEED in any direction?
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).Faiq said:I know the velocity is changing because of the direction. 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 can't we apply the same analogy in circular motion? More importantly, why isn't force giving rise to a component of SPEED in any direction?
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 traveled a small distance in the direction of force.sophiecentaur said: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.
So?Faiq said: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 traveled a small distance in the direction of force.
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?jbriggs444 said: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.
There will be a gain in energy which could compensate for the increase in kinetic energy.Chestermiller said:So?
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,Faiq said: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 traveled a small distance in the direction of force.
You are basically asking how the Calculus method can be used to establish the v2/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.Faiq said: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?
How can there be any change in the Energy if the motion is not speeding up or slowing down?Faiq said:there could be an increase in energy.
I can comprehend calculus but calculus wasn't used to derive the formula v2/r (when I was taught)sophiecentaur said:You are basically asking how the Calculus method can be used to establish the v2/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.
No. It does not. Not at all. It travels now in the direction of the force a moment ago, but that is not the same thing.Faiq said:But the object does travel in the direction of forc
How can it, if the string doesn't get shorter?Faiq said:But the object does travel in the direction of force
How can you say that calculus was not used? Your last post was full of it.Faiq said:I can comprehend calculus but calculus wasn't used to derive the formula v2/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 = v2/r
A force with a constant direction is not centripetal.Faiq said:Okay I have a thought experiment. Let's say an object A and B are traveling at constant speeds. Both objects are subjected to a same force for a very small time. The force is centripetal in nature to A but is just a normal resultant force to B. So after that time will both objects travel at different velocities, as the A's velocity will remain unchanged? Why is that contradiction arising?
The direction is nearly constant. But that nearly is exactly the problem. The resulting speed is nearly unchanged by a constant force. If you account for the fact that the direction is changing then the resulting speed is exactly unchanged.Faiq said:Direction for A is not constant, but the time is very small
What is the difference between the motions? At any instant, the object moving in a circle would travel in a straight tangential line if the force were removed (there is no 'memory' of what it's doing. A and B would be deflected by exactly the same amount.Faiq said:The force is centripetal in nature to A but is just a normal resultant force to B.
Yes that's what I am trying to askChestermiller said:Faiq:
Please correct me if I am wrong.
I think what you are asking is the following: If a mass is traveling in a horizontal circle on a frictionless table, tethered by a string at the center of the circle, is a constant speed for the mass consistent with the Newton's 2nd law force balances on the mass in the x and y directions (Cartesian coordinates) for the entire motion around a circle? In other words, is there as solution to Newton's 2nd law equations consistent with constant speed motion of the mass in a circle tethered at the center with a string (even if one uses Cartesian coordinates to analyze the motion)?
Does this capture the essence of what you are asking?
There is no such thing as a 'next' instant. Time is continuous, not discrete. The acceleration is only vertical for an instant of zero duration.Faiq said:Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with Newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?
Why don't you use the equations I presented to help you analyze this exact situation?Faiq said:Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with Newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?
no.Faiq said:Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with Newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?