Can an object maintain uniform motion without any external force?

[Silverbeam]

It is not in uniform motion, it is in uniform circular motion. Those are different. The first and second velocities will be different.

Wouldn't that mean it is speeding up though? Covering more distance in less time?

 

[Dale]

Wouldn't that mean it is speeding up though? Covering more distance in less time?

No. It is turning. Do you understand the difference between speed and velocity? Perhaps that is the problem.

 

[DrStupid]

Could you explain what you mean by this?

The position of an object in uniform circular motion is

$r = r_0 + R \cdot \left( {\begin{array}{*{20}c} {\cos \left( {\varphi _0 + \omega \cdot t} \right)} \\ {\sin \left( {\varphi _0 + \omega \cdot t} \right)} \\ \end{array}} \right)$

the velocity is

$\dot r = \omega \cdot R \cdot \left( {\begin{array}{*{20}c} { - \sin \left( {\varphi _0 + \omega \cdot t} \right)} \\ {\cos \left( {\varphi _0 + \omega \cdot t} \right)} \\ \end{array}} \right)$

and the acceleration

$\ddot r = \omega ^2 \cdot R \cdot \left( {\begin{array}{*{20}c} { - \cos \left( {\varphi _0 + \omega \cdot t} \right)} \\ { - \sin \left( {\varphi _0 + \omega \cdot t} \right)} \\ \end{array}} \right)$

 

[Silverbeam]

No. It is turning. Do you understand the difference between speed and velocity? Perhaps that is the problem.

Perhaps. I thought an object in uniform circular motion went at a constant speed, so that it would always take the same amount of time to travel indentical portions of the circle.

 

[Dale]

Perhaps. I thought an object in uniform circular motion went at a constant speed, so that it would always take the same amount of time to travel indentical portions of the circle.

That is correct. It goes at a constant speed. Now, what about the velocity, is that constant?

 

[davenn]

Perhaps. I thought an object in uniform circular motion went at a constant speed, so that it would always take the same amount of time to travel indentical portions of the circle.

That didnt answer Dale's Q to you

do you understand the difference between speed and velocity ?

 

[PeroK]

Wouldn't that mean it is speeding up though? Covering more distance in less time?

Really, you need to take some time to learn a bit of physics:

https://www.khanacademy.org/science...rcular-motion-ap/a/circular-motion-basics-ap1

 

[Silverbeam]

No. It is turning. Do you understand the difference between speed and velocity? Perhaps that is the problem.

Sorry, I didn't answer your question. No, I don't understand the difference between speed and velocity.

 

[DrStupid]

For the ISS, the two accelerations are aligned and subtract to zero.

That's what I'm talking about. The ISS is accelerated in classical physics but the accelerometer shows zero. It doesn't work this way.

 

[Dale]

Sorry, I didn't answer your question. No, I don't understand the diffrence between speed and velocity.

Velocity is what is called a vector. That means that it has a magnitude and a direction. You can think of it as an arrow. The magnitude is its length and its direction is the direction it points.

For velocity, the magnitude is called speed. So the speed is what is left of velocity when you remove the direction. But acceleration is a change in velocity, not a change in speed. So to determine acceleration you have to keep the direction information, not just the magnitude information. That is what is changing in uniform circular motion.

 

[Silverbeam]

Velocity is what is called a vector. That means that it has a magnitude and a direction. You can think of it as an arrow. The magnitude is its length and its direction is the direction it points.

For velocity, the magnitude is called speed. So the speed is what is left of velocity when you remove the direction. But acceleration is a change in velocity, not a change in speed. So to determine acceleration you have to keep the direction information, not just the magnitude information. That is what is changing in uniform circular motion.

Ok, thanks.

 

[etotheipi]

No, it does work for both the car and the ISS. The phone has a 3-axis accelerometer, and the acceleration measured when stationary on the surface of the Earth is 1.0g vertically. The acceleration from the car turning is measured separately, in a perpendicular axis. For the ISS, the two accelerations are aligned and subtract to zero.

I don't really know what you mean by 'the two accelerations are aligned and subtract to zero' for the ISS case. When we're thinking in terms of GR and all that stuff, if we ignore the atmosphere then there's no force at all on the ISS (apart from the tidal forces, which we can ignore on the small scales), and so no proper acceleration. Rather than 'two accelerations cancelling', probably clearer to say that there was no proper acceleration in the first place.

 

[Silverbeam]

Velocity is what is called a vector. That means that it has a magnitude and a direction. You can think of it as an arrow. The magnitude is its length and its direction is the direction it points.

For velocity, the magnitude is called speed. So the speed is what is left of velocity when you remove the direction. But acceleration is a change in velocity, not a change in speed. So to determine acceleration you have to keep the direction information, not just the magnitude information. That is what is changing in uniform circular motion.

So, if I want to approach a corner in my car, driving in a straight line, then take the corner, then drive straight ahead, all at a constant speed, do I need to provide any additional force to the car while taking the corner in order to maintain the constant speed throughout?

 

[etotheipi]

So, if I want to approach a corner in my car, driving in a straight line, then take the corner, then drive straight ahead, all at a constant speed, do I need to provide any additional force to the car while taking the corner in order to maintain the constant speed throughout?

Friction from the road surface provides the necessary inward centripetal component of force. also of course this is an idealisation

In general when a body moves along a trajectory with radius of curvature $\rho$ at constant speed $v$, there is acceleration of magnitude $v^2/\rho$ directed toward the instantaneous centre of curvature, which must be caused by a net force also directed toward the instantaneous centre of curvature.

(n.b. that in most general case where $\dot{v} \neq 0$, you will obtain an additional component of acceleration $\dot{v} \boldsymbol{e}_t$ as well as the inward component $(v^2 / \rho) \boldsymbol{e}_n$)

 

[Silverbeam]

friction from the road surface provides the necessary inward centripetal component of force. also of course this is an idealisation

in general when a body moves along a trajectory with radius of curvature $\rho$ at constant speed $v$, there is acceleration of magnitude $v^2/\rho$ directed toward the instantaneous centre of curvature, which must be caused by a net force also directed toward the instantaneous centre of curvature

So, I would not need to apply any more acceleration to the vehicle while driving in identical circles at a constant speed, than the acceleration I would need to apply if I were driving in a straight line at a constant speed?

 

[etotheipi]

No that's not what i said at all! a body traveling in a straight line at constant speed undergoes zero acceleration. n.b. that this is consistent with the relation i wrote, because a straight line has radius of curvature $\rightarrow \infty$.

 

[Dale]

So, if I want to approach a corner in my car, driving in a straight line, then take the corner, then drive straight ahead, all at a constant speed, do I need to provide any additional force to the car while taking the corner in order to maintain the constant speed throughout?

For a car, yes, but for a marble on a smooth track that turns, no. The energy required for the car has to do with inefficiency, not fundamental physical principles.

 

[Silverbeam]

For a car, yes, but for a marble on a smooth track that turns, no. The energy required for the car has to do with inefficiency, not fundamental physical principles.

Ok, but for a marble on a track it has the sides of the track to keep it in a circle. What about a marble going in circles without a track? Doesn't it need some force to be applied to it, either attractive or repulsive, in order to constantly change its direction and keep it in a circle? Isn't that why gravity is invoked for orbits, to provide a force that maintains the inward acceleration to prevent the Moon or whatever from just going off in a straight line?

 

[etotheipi]

Ok, but for a marble on a track it has the sides of the track to keep it in a circle. What about a marble going in circles without a track? Doesn't it need some force to be applied to it, either attractive or repulsive, in order to constantly change its direction and keep it in a circle? Isn't that why gravity is invoked for orbits, to provide a force that maintains the inward acceleration to prevent the Moon or whatever from just going off in a straight line?

yes!

 

[Dale]

Doesn't it need some force to be applied to it, either attractive or repulsive, in order to constantly change its direction and keep it in a circle?

Yes, it does.

I may have misunderstood your question earlier. I thought you were asking if you would need to step on the gas in a car to maintain speed through a turn. It is not necessary for a marble to “step on the gas” as it were. It needs the track, but not a push along the track, unlike a car which needs the push too.

Isn't that why gravity is invoked for orbits, to provide a force that maintains the inward acceleration to prevent the Moon or whatever from just going off in a straight line?

Yes.

 

[Silverbeam]

yes!

But Dale says the only additional force I need to turn a corner in my car with constant speed is due to inefficiency, not a fundamental physical principle. Why doesn't the same additional force need to be required for the car as for the Moon?

 

[Silverbeam]

I may have misunderstood your question earlier. I thought you were asking if you would need to step on the gas in a car to maintain speed through a turn. It is not necessary for a marble to “step on the gas” as it were.

That is what I meant.

 

[etotheipi]

But Dale says the only additional force I need to turn a corner in my car with constant speed is due to inefficiency, not a fundamental physical principle. Why doesn't the same additional force need to be required for the car as for the Moon?

sorry, I don't understand what you mean about inefficiency and fundamental physics principles

if the centre of mass of the car is accelerating at some $\ddot{\mathbf{x}} = (v^2 / \rho) \mathbf{e}_n$, then the net force on the car must be $m\ddot{\mathbf{x}} = (mv^2 / \rho)\mathbf{e}_n$. the only external force on the car in such a situation, which can provide that force, is friction.

 

[Dale]

Ok, let’s start with the basics.

You now understand the difference between velocity and speed. Acceleration is a change in velocity. Acceleration, like velocity, is a vector so it has a direction as well as a magnitude. When acceleration is parallel to the velocity then it changes the speed. When acceleration is perpendicular to the velocity then it changes the direction of the velocity.

Force is proportional to acceleration. So in order to go in circular motion you need a force that is perpendicular to the velocity. That force is required in order to turn.

When a marble turns on a smooth track the force is perpendicular to the velocity. So a marble can turn without any additional push from behind. It only needs the push from the side given by the track.

When a car turns, because of how tires work, the force is not perfectly perpendicular to the velocity. There is a little bit of force going anti-parallel to the velocity. So you need a touch of gas to counteract that. You need the push from the side but also a bit of a push from behind due to the inefficiency of turning with tires.

 

[etotheipi]

When a car turns, because of how tires work, the force is not perfectly perpendicular to the velocity. There is a little bit of force going anti-parallel to the velocity. So you need a touch of gas to counteract that. You need the push from the side but also a bit of a push from behind due to the inefficiency of turning with tires.

I think the key point to understand here is that no matter how complex the many different forces between the tyres and the road, the sum of all these forces from the road as well as of course the weight (which, neglecting the air, are the only net external forces on the car) is always proportional [parallel] to the acceleration of the centre of the mass of the car

 

[hmmm27]

If you take a pinball and shoot it directly at (the inside of) a V, it will bounce off one arm, go sideways, bounce off the other, then run back where it came from, experiencing no loss of speed at any time. If you were inside the pinball, you'd get creamed at each bounce because of the massive instantaneous change in velocity.

Now shoot it at a U ; it slides down the inside, curves around the bottom, then procedes to roll back in the opposite direction. If you were inside that one you'd be much more comfortable, since the changing velocity (ie: acceleration) is spread out over time. Like the V example, the speed doesn't change throughout the procedure.

 

[Dale]

I think the key point to understand here is that no matter how complex the many different forces between the tyres and the road, the sum of all these forces from the road as well as of course the weight (which, neglecting the air, are the only net external forces on the car) is always proportional [parallel] to the acceleration of the centre of the mass of the car

Yes agreed.

I only explain about tires because if they actually do the experiment they will find that they do slow down. That isn’t because the theory is wrong, but because cars are complicated and messy machines. The marbles are simpler and easier to analyze. I always prefer simple experiments.

 

[etotheipi]

but because cars are complicated and messy machines. The marbles are simpler and easier to analyze. I always prefer simple experiments.

Yes, exactly! Much better to build up complexity gradually, rather than starting with an extremely complex mechanism like a car! There is much to be gained by carefully understanding the dynamics of a single particle, then generalising to a system of particles, then a system of interacting particles, continuous bodies etc.

e.g. your example of the marble is a much better tool to understand the general motion of a particle than the example with the car that the op kept referring to

 

[russ_watters]

That's what I'm talking about. The ISS is accelerated in classical physics but the accelerometer shows zero. It doesn't work this way.

"Shows zero" and "sums to zero" is the same thing here. I'll explain:

I don't really know what you mean by 'the two accelerations are aligned and subtract to zero' for the ISS case. When we're thinking in terms of GR and all that stuff, if we ignore the atmosphere then there's no force at all on the ISS (apart from the tidal forces, which we can ignore on the small scales), and so no proper acceleration. Rather than 'two accelerations cancelling', probably clearer to say that there was no proper acceleration in the first place.

• A stationary or constant speed car with a 3-axis accelerometer reads 1.0g upward acceleration.
• This car has excellent tires. It turns a hard circle to the left, at 1g. The accelerometer reads 1.0g up and 1.0g to the left, for a sum of 1.4g, at a 45 degree angle from vertical.
• Now the car does the same turn on a 45 degree banked track. To be clear: I mean the same turn, in the plane of the turn, not in the horizontal (Earth's surface) plane. Now we have the upwards acceleration of 1g and a downward and to the left acceleration of 1g at the top of the turn; add them together and you get a resultant 0.7g. Note: the accelerometer doesn't read them as 1g in each axis here because they are starting to oppose each other.
• Now the car does the same "turn" in a loop. At the top of the loop, the accelerometer reads...

 

[A.T.]

• This car has excellent tires. It turns a hard circle to the left, at 1g. The accelerometer reads 1.0g up and 1.0g to the left, for a sum of 1.4g, at a 45 degree angle from vertical.
• Now the car does the same turn on a 45 degree banked track. To be clear: I mean the same turn, in the plane of the turn, not in the horizontal (Earth's surface) plane. Now we have the upwards acceleration of 1g and a downward and to the left acceleration of 1g at the top of the turn; add them together and you get a resultant 0.7g. Note: the accelerometer doesn't read them as 1g in each axis here because they are starting to oppose each other.

A banked turn just means that there is less/none lateral tire friction needed, because the normal force has a centripetal component. If the trajectory of the car is still the same as for the flat track, the proper acceleration will be of the same magnitude.

Or do you mean the entire circular flat track is now on an inclined plane?