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.
–
Sai Prasanth Rao
·
2 years, 4 months ago

Log in to reply

@Sai Prasanth Rao
–
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 \(0\times \infty \)
–
Abhineet Nayyar
·
2 years, 4 months ago

Log in to reply

@Abhineet Nayyar
–
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
·
2 years, 4 months ago

Log in to reply

@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
·
2 years, 4 months ago

Log in to reply

@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
·
2 years, 4 months ago

Log in to reply

@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...
–
Abhineet Nayyar
·
2 years, 4 months ago

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...:)
–
Abhineet Nayyar
·
2 years, 4 months ago

Log in to reply

yes if there are infinite number of 9's
–
Sai Prasanth Rao
·
2 years, 4 months ago

Log in to reply

Can \(1\) be thought of as \(0.99999999999999999.........................................\) ?
–
Satvik Golechha
·
2 years, 4 months ago

## 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. – Sai Prasanth Rao · 2 years, 4 months ago

Log in to reply

– Abhineet Nayyar · 2 years, 4 months ago

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

– Sai Prasanth Rao · 2 years, 4 months ago

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

– Abhineet Nayyar · 2 years, 4 months ago

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 motionLog in to reply

– Sai Prasanth Rao · 2 years, 4 months ago

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)rLog in to reply

– Abhineet Nayyar · 2 years, 4 months ago

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...Log in to reply

– Sai Prasanth Rao · 2 years, 4 months ago

which standard are you in?Log in to reply

– Abhineet Nayyar · 2 years, 4 months ago

11, is there anything wrong?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...:) – Abhineet Nayyar · 2 years, 4 months ago

Log in to reply

yes if there are infinite number of 9's – Sai Prasanth Rao · 2 years, 4 months ago

Log in to reply

Can \(1\) be thought of as \(0.99999999999999999.........................................\) ? – Satvik Golechha · 2 years, 4 months ago

Log in to reply