Can rectilinear motion be thought of as a circular motion with radius of curvature that tends to infinity.

Note by Sai Prasanth Rao
5 years, 2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Can 11 be thought of as 0.99999999999999999.........................................0.99999999999999999......................................... ?

Satvik Golechha - 5 years, 2 months ago

Log in to reply

yes if there are infinite number of 9's

Sai Prasanth Rao - 5 years, 2 months ago

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=ωrv=\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.....0.9999999.....\infty tends to one, but considering it as one is an approximation, i.e. 0.99999....10.99999....\infty \approx 1, not an equality...:)

Abhineet Nayyar - 5 years, 2 months ago

Log in to reply

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 - 5 years, 2 months ago

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 0×0\times \infty

Abhineet Nayyar - 5 years, 2 months ago

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.

Sai Prasanth Rao - 5 years, 2 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 mv2r\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 - 5 years, 2 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 - 5 years, 2 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 - 5 years, 2 months ago

Log in to reply

@Abhineet Nayyar which standard are you in?

Sai Prasanth Rao - 5 years, 2 months ago

Log in to reply

@Sai Prasanth Rao 11, is there anything wrong?

Abhineet Nayyar - 5 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...