When you say 0.99999.. with a large number of 9's it would tend to 1 but when you say infinite number of 9's it should be equal to 1 as it is an infinite G.P. As far as the note is concerned, a circle with infinite radius would give an object zero angular velocity. Another example-A polygon with infinite sides is a circle.

You said it yourself...zero angular velocity and an infinite radius, how will you define their product i.e. Linear velocity, because you can't define \(0\times \infty \)

however we can define the formula (m)(v^2)/r. As r is infinity the accelaration is zero. However how is it possible that in this way we get zero centripetal accelaration and in your method we get undefined centripetal accelaration.

@Sai Prasanth Rao
–
In my way, you don't get undefined centripetal acceleration, Although you get an undefined linear velocity.
And moreover using \(\frac { m{ v }^{ 2 } }{ r } \) , we see that an object moving in a circle must have a centripetal force. But, in this case, the centripetal force is zero, which clearly signifies that the object is not in circular motion

@Abhineet Nayyar
–
yes we can conclude that the object is not in circular motion if we use the formula mv^2/r, but why can't we conclude the result while using the formula (mw^2)r

@Sai Prasanth Rao
–
Yes you can, as you can see 2 formulas of the same quantity are condradicting each other. That means that our earlier assumption that an object moving linearly is a part of a circle with an infinite radius is wrong...

As far as i think rectilinear motion can not be considered as circular motion, because with an infinite radius and using the formula \(v=\omega r\), even for a little bit of an angular velocity, our linear velocity should turn out be infinite which is not true, as it is possible to measure a speed of a moving car but not of infinity...

Moreover \(0.9999999.....\infty \) tends to one, but considering it as one is an approximation, i.e. \(0.99999....\infty \approx 1\), not an equality...:)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWhen you say 0.99999.. with a large number of 9's it would tend to 1 but when you say infinite number of 9's it should be equal to 1 as it is an infinite G.P. As far as the note is concerned, a circle with infinite radius would give an object zero angular velocity. Another example-A polygon with infinite sides is a circle.

Log in to reply

You said it yourself...zero angular velocity and an infinite radius, how will you define their product i.e. Linear velocity, because you can't define \(0\times \infty \)

Log in to reply

however we can define the formula (m)(v^2)/r. As r is infinity the accelaration is zero. However how is it possible that in this way we get zero centripetal accelaration and in your method we get undefined centripetal accelaration.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

As far as i think rectilinear motion can not be considered as circular motion, because with an infinite radius and using the formula \(v=\omega r\), even for a little bit of an angular velocity, our linear velocity should turn out be infinite which is not true, as it is possible to measure a speed of a moving car but not of infinity...

Moreover \(0.9999999.....\infty \) tends to one, but considering it as one is an approximation, i.e. \(0.99999....\infty \approx 1\), not an equality...:)

Log in to reply

yes if there are infinite number of 9's

Log in to reply

Can \(1\) be thought of as \(0.99999999999999999.........................................\) ?

Log in to reply