Angular accn

A rod of length ll 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 aa and bb respectively.

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

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

This means x¨=a=lcosθθ˙2lsinθθ¨\ddot{x}=a=-l\cos\theta\dot{\theta}^2-l\sin\theta\ddot{\theta} and y¨=b=lsinθθ˙2+lcosθθ¨\ddot{y}=b=-l\sin\theta\dot{\theta}^2+l\cos\theta\ddot{\theta} and squaring and adding gives me a2+b2l2=θ˙4+θ¨2\dfrac{a^2+b^2}{l^2}=\dot{\theta}^4+\ddot{\theta}^2.

But , The correct answer is a2+b2l2=θ¨2\dfrac{a^2+b^2}{l^2}=\ddot{\theta}^2

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

Note by Vilakshan Gupta
11 months, 1 week 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

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

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

Vilakshan Gupta - 11 months, 1 week ago

Log in to reply

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.

Steven Chase - 11 months, 1 week ago

Log in to reply

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

Aaghaz Mahajan - 11 months, 1 week ago

Log in to reply

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

Ram Mohith - 11 months, 1 week ago

Log in to reply

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???

Aaghaz Mahajan - 11 months, 1 week ago

Log in to reply

@Aaghaz Mahajan u come on slack now...

Vilakshan Gupta - 11 months, 1 week ago

Log in to reply

@Aaghaz Mahajan I am in grade 11 from Visakhapatnam, andra Pradesh.

Ram Mohith - 11 months, 1 week ago

Log in to reply

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 θ\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).

Mark Hennings - 11 months, 1 week ago

Log in to reply

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...

Vilakshan Gupta - 11 months, 1 week ago

Log in to reply

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 00. Thus θ˙4+θ¨2  =  θ¨2 \dot{\theta}^4 + \ddot{\theta}^2 \; = \; \ddot{\theta}^2 initially.

Mark Hennings - 11 months, 1 week ago

Log in to reply

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

Vilakshan Gupta - 11 months, 1 week ago

Log in to reply

@Vilakshan Gupta 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.

Mark Hennings - 11 months, 1 week ago

Log in to reply

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

Vilakshan Gupta - 11 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...