# Angular accn

A rod of length $l$ is sliding such that one of it's ends is always in contact with a vertical wall and it's other end is always in contact with horizontal surface ( rod is lying in vertical plane ). Just after the rod is released from rest, the magnitude of acceleration of end points of the rod is $a$ and $b$ respectively.

What will be the angular acceleration of the rod at this instant?

My attempt: If we assume that the rod makes an angle $\theta$ with $x$ axis, then $x=l\cos\theta$ and $y=l\sin\theta$ which means $\dot{x}=-l\sin\theta\dot{\theta}$ and $\dot{y}=l\cos\theta\dot{\theta}$ (Here $\dot{\theta}<0$)

This means $\ddot{x}=a=-l\cos\theta\dot{\theta}^2-l\sin\theta\ddot{\theta}$ and $\ddot{y}=b=-l\sin\theta\dot{\theta}^2+l\cos\theta\ddot{\theta}$ and squaring and adding gives me $\dfrac{a^2+b^2}{l^2}=\dot{\theta}^4+\ddot{\theta}^2$.

But , The correct answer is $\dfrac{a^2+b^2}{l^2}=\ddot{\theta}^2$

So does that mean, just after releasing, $\dot{\theta}=0?$

Note by Vilakshan Gupta
2 years, 8 months ago

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.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Have a look at this problem. We are assuming (I presume) that the contact forces with the wall and floor are smooth.

If normal reaction forces have to be positive, the rod does not stay in contact with the wall forever, but the differential equation I give for $\theta$ can be differentiated to give the initial angular acceleration, and therefore also the initial accelerations of the endpoints. If the rod is to stay in contact with the wall throughout the motion, the reaction with the wall will have be be allowed to be negative (so the rod is sliding down a smooth groove in the wall, of the like).

- 2 years, 8 months ago

but I am asking what I have done wrong? My case deals only with an immediate instant after the release of the rod and not with throughout motion of the rod... rod doesn't know that it will be leaving the contact with the wall later in the motion.... I have also written some extra calculation how I did it...

- 2 years, 8 months ago

You haven't done anything wrong, except forget the phrased "released from rest". The rod starts at rest, and so its initial angular speed is $0$. Thus $\dot{\theta}^4 + \ddot{\theta}^2 \; = \; \ddot{\theta}^2$ initially.

- 2 years, 8 months ago

Okay...but can you tell whether we can do it by Instantaneous centre/axis of rotation method?

- 2 years, 8 months ago

Well, since the rod is stationary initially, all points are instantaneously at rest, and so you can take any point as the instantaneous centre of rotation.

- 2 years, 8 months ago

Also I wanted to mention without differentiation... any elegant method...because differentiation is not that much elegant....

- 2 years, 8 months ago

@Steven Chase @Mark Hennings Sir where I am wrong?

And also tell how can we solve it by Instantaneous Centre of Rotation method.

- 2 years, 8 months ago

I don't typically use that method. But from what I've read, it sounds like the instantaneous center method is useful for relating the two endpoint accelerations, a and b, in a simple way.

- 2 years, 8 months ago

@Vilakshan Gupta Come on slack............

- 2 years, 8 months ago

Can you tell me how to join slack. I am trying for it for many days.

- 2 years, 8 months ago

Yes surely............see.......earlier, there was a Brilliant Lounge on slack.......but it has been deactivated due to lack of activity on it.........Now, what me and Vilakshan Gupta are doing is that, we have joined an online community for RMO/INMO preparation, and then we use personal chats to talk..........that can work........Should I send you an invite???
Btw, Which class are you in?? And where are you from???

- 2 years, 8 months ago

u come on slack now...

- 2 years, 8 months ago