# Velocities in circular motion

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?

Chestermiller
Mentor
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.

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.

Khashishi
That's what the calculus tells you. If you kept accelerating in the same direction, then you would change the speed. But in a circular motion, you constantly change the direction of the acceleration. If you consider an infinitesimal time-step of size epsilon, the amount of speed change is proportional to epsilon squared. So, in the limit of epsilon->0, there is no speed change.

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

"Wouldn't the force give rise to a vertical component of velocity on the object"- It would, but without changing the magnitude of the velocity vector, and thus only changes the direction of velocity.
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.

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 cant we apply the same analogy in circular motion? More importantly, why isn't force giving rise to a component of SPEED in any direction?

A.T.
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?
Because you cannot apply a force perpendicularly to a velocity of zero.

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

sophiecentaur
Gold Member
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 cant we apply the same analogy in circular motion? 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.
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.

jbriggs444
Homework Helper
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 cant 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).

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.

Chestermiller
Mentor
It is not changing speed because at no time is there a component of force in the direction that it is moving.

jbriggs444 and nasu
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.
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.

Chestermiller
Mentor
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.
So???

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.
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?

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

sophiecentaur
Gold Member
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.
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,

sophiecentaur
Gold Member
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 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.

But the object does travel in the direction of force so according to Energy = Force * Displacement there could be an increase in energy. Btw that wasn't my question.
My question is simply why is the magnitude of velocity not changing even though a force is acting on the body?
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

sophiecentaur
Gold Member
there could be an increase in energy.
How can there be any change in the Energy if the motion is not speeding up or slowing down?
[or vice versa].

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.
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

jbriggs444
Homework Helper
But the object does travel in the direction of forc
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.

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

sophiecentaur
Gold Member
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
How can you say that calculus was not used? Your last post was full of it.

Okay I have a thought experiment. Lets say an object A and B are travelling 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?

jbriggs444
Homework Helper
Okay I have a thought experiment. Lets say an object A and B are travelling 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?
A force with a constant direction is not centripetal.

Direction for A is not constant, but the time is very small

jbriggs444
Homework Helper
Direction for A is not constant, but the time is very small
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.

sophiecentaur
Chestermiller
Mentor
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?

sophiecentaur
sophiecentaur
Gold Member
The force is centripetal in nature to A but is just a normal resultant force to B.
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.
You need to understand how the calculus can be used in situations like this. You have to go through the derivation with extreme care in order to work out the limits for the quantities are as δt → 0 when you start the calculation using finite time and small straight line motion. Some quantities are unchanged and some go to zero (as with all differentiation from first principles when you start with tan(θ) = δy/δx etc. etc.) You quoted dx/dt but how was it derived from scratch? Understand that and your questions and worries will be answered. It was second year A level for me, way back and Mr. Worthington was extremely meticulous about it all.

Chestermiller
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?
Yes that's what I am trying to ask

Chestermiller
Mentor
OK. Let's see how this plays out. The equation for a circle is $$x^2+y^2=r^2\tag{1}$$. The circular motion of the mass moving at constant speed can be expressed parametrically as $$x=r\cos \left(\frac{v}{r}t\right)\tag{2}$$
$$y=r\sin \left(\frac{v}{r}t\right)\tag{3}$$where v is the (constant) speed and t is time. Eqns. 2 and 3 satisfy Eqn. 1 exactly. How do we know that this is also consistent with a constant speed for the mass? If we take the derivative of the coordinates of the mass x and y with respect to time, we obtain the x and y components of velocity:
$$v_x=-v\sin \left(\frac{v}{r}t\right)\tag{4}$$
$$v_y=+v\cos \left(\frac{v}{r}t\right)\tag{5}$$
If we take the sum of the squares of these two velocity components, we (automatically) obtain the square of the (constant) speed v:
$$v_x^2+v_y^2=v^2\tag{6}$$
Not let T be the tension of the string attached to the origin. At any time t, the components of the tension in the x and y directions are:
$$T_x=-T\frac{x}{\sqrt{x^2+y^2}}=-T\cos \left(\frac{v}{r}t\right)\tag{7}$$
$$T_y=-T\frac{y}{\sqrt{x^2+y^2}}=-T\sin \left(\frac{v}{r}t\right)\tag{8}$$
Now for the force balances. The force balance in the x direction is given by:
$$m\frac{dv_x}{dt}=T_x=-T\cos \left(\frac{v}{r}t\right)\tag{9}$$
The force balance in the y direction is given by:$$m\frac{dv_y}{dt}=T_y=-T\sin \left(\frac{v}{r}t\right)\tag{10}$$

@Faiq: Is this all OK with you so far?

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?

jbriggs444
Homework Helper
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?
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.

Edit to add: If you look at the object after it has attained some non-zero vertical velocity then the average acceleration over the non-zero duration that it took to acquire that velocity will not have been purely vertical.

Chestermiller
Mentor
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?

sophiecentaur