Two points of contact in rolling

[Kashmir]

Suppose a sphere is rolling on horizontal surface. The point of contact is the instantaneous center of rotation. It's velocity momentarily is zero. So in a time dt it'll stay where it is.
However during this time dt the point next to this contact point at its right side will move forward and down, thus after time dt we will have two points in contact with the horizontal surface. Which is wrong.

What's wrong with my approach?

 

[Ibix]

What's wrong with my approach?

You are allowing the positions of the points to change in time dt, but not their velocities. That means you are modelling their instantaneous accelerations as zero, which is not consistent with a model of a rotating object.

 

[PeroK]

More fundamentally, $dt$ is either a non-zero time interval (in which case nothing is constant) or it's a differential (in which case it is not a time interval).

 

[Kashmir]

You are allowing the positions of the points to change in time dt, but not their velocities. That means you are modelling their instantaneous accelerations as zero, which is not consistent with a model of a rotating object.

If the point of contact has zero velocity it means in a time interval dt $dr=0$ because $dr=v.dt$.
But I didn't say that the velocity didn't change in dt, after dt the contact point has a non zero velocity.

 

[Kashmir]

More fundamentally, $dt$ is either a non-zero time interval (in which case nothing is constant) or it's a differential (in which case it is not a time interval).

By dt I mean the thing we mean when writing $dr=v dt$

 

[PeroK]

By dt I mean the thing we mean when writing $dr=v dt$

That's a differential (a mathematical concept); and not a time interval.

 

[Kashmir]

That's a differential (a mathematical concept); and not a time interval.

So all that introductory books that say "the small displacement $dr=v.dt$ in time interval $dt$ are simplifications ?
All the introductory mechanics never talks about anything else. But when I ask this question then we need differentials?

 

[PeroK]

So all that introductory books that say "the small displacement $dr=v.dt$ in time interval $dt$ are simplifications ?

They are technically wrong. They should use $\Delta r = v\Delta t$ for a small displacement.

All the introductory mechanics never talks about anything else. But when I ask this question then we need differentials?

If you use $dt$ as a smal time interval, then it's a small time interval. You can't then also assume it is zero.

 

[sophiecentaur]

So all that introductory books that say "the small displacement $dr=v.dt$ in time interval $dt$ are simplifications ?
All the introductory mechanics never talks about anything else. But when I ask this question then we need differentials?

There's a very subtle Mathematician's step that's often ignored by 'non-mathematicians' who use Maths. The strict definition of a differential is 'the limit of δy/δx ( δy and δx are small but finite quantities - say the steps on graph) as δx approaches zero. Why use a limit? Because you can often fall into the trap of dividing zero by zero, which has no meaning. But using the limiting value of dy/dx allows you to do the apparent division safely and keeps everything sanitary. Your rotation question needs this treatment if you want the 'right' answer.

My hero Mr Worthington started us on differential calculus and made us do it properly. I am eternally thankful for how he taught us in the early 1960s. No messing'.
(OTOH, you can just accept it and not lose sleep over it.)

 

[Herman Trivilino]

Suppose a sphere is rolling on horizontal surface. The point of contact is the instantaneous center of rotation. It's velocity momentarily is zero. So in a time dt it'll stay where it is.

No, it won't. It will, for example, rotate through an angle $d\theta$.

 

[Ibix]

If the point of contact has zero velocity it means in a time interval dt $dr=0$ because $dr=v.dt$.
But I didn't say that the velocity didn't change in dt, after dt the contact point has a non zero velocity.

But $dr$ is only zero if $v$ is zero for the whole time period $dt$. So you are saying that the acceleration is zero - which I hope you know is false. If the contact point has non-zero velocity after the time period, how did it get that way if the acceleration is zero?

The point is that if you are using $dt$ to mean a small time interval then $v$ is actually $v(t)$ and not a constant. If, instead, you are meaning that $v$ is the instantaneous velocity then $dt$ is not a time period. Rather, it is the limit as the time period goes to zero, which is a mathematical device that allows you to deal with an instant of zero duration without having to divide by zero, as others have explained in more detail.

 

[sophiecentaur]

But $dr$ is only zero if $v$ is zero for the whole time period $dt$. So you are saying that the acceleration is zero - which I hope you know is false. If the contact point has non-zero velocity after the time period, how did it get that way if the acceleration is zero?

The point is that if you are using $dt$ to mean a small time interval then $v$ is actually $v(t)$ and not a constant. If, instead, you are meaning that $v$ is the instantaneous velocity then $dt$ is not a time period. Rather, it is the limit as the time period goes to zero, which is a mathematical device that allows you to deal with an instant of zero duration without having to divide by zero, as others have explained in more detail.

Yes. It's important to distinguish between your 'd's and your 'deltas' if you want a bombproof explanation. In differential calculus, dt is an indeterminate amount even when dx/dt has a value. I think that arm waving and calculus don't work together well if you are trying to have a rigorous argument. Having said that, the days are long gone when I had it all sorted out formally in my head. All I really remember is that you gotta be very careful.