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, $\dot{\theta}=0?$

#Mechanics

Easy Math Editor

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Comments

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).

Mark Hennings - 2 years, 7 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...

Vilakshan Gupta - 2 years, 7 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.

Mark Hennings - 2 years, 7 months ago

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

Vilakshan Gupta - 2 years, 7 months ago

@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 - 2 years, 7 months ago

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

Vilakshan Gupta - 2 years, 7 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.

Vilakshan Gupta - 2 years, 7 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.

Steven Chase - 2 years, 7 months ago

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

Aaghaz Mahajan - 2 years, 7 months ago

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

Ram Mohith - 2 years, 7 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???

Aaghaz Mahajan - 2 years, 7 months ago

@Aaghaz Mahajan – u come on slack now...

Vilakshan Gupta - 2 years, 7 months ago

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

Ram Mohith - 2 years, 7 months ago

Math	Appears 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}`	$\boxed{123}$