What Is the Velocity of End B When the Rod Makes a 60° Angle with the Ground?

AI Thread Summary
The discussion revolves around calculating the velocity of end B of a rod when it makes a 60° angle with the ground, given that end A is fixed and a block moves with a velocity of √3 m/s. The user initially derived the wrong vertical velocity of end B, calculating it as -3 m/s instead of the correct answer of 2 m/s. After re-evaluating their calculations, they realized a mistake in applying the tangent function for the angle. Once corrected, they confirmed that the vertical component of B's velocity is indeed -1 m/s, leading to the conclusion that the overall velocity of end B is 2 m/s. The conversation highlights the importance of careful mathematical verification in physics problems.
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Homework Statement


A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity √3 m/s towards right. The velocity of end B when rod makes an angle of 60o with the ground.
90yf4g.jpg




Homework Equations





The Attempt at a Solution


I assumed that at any instant the distance of A from block is x and distance of B from ground is y. The length of rod is l.
l^2=x^2+y^2
Differentiating with respect to time.
0=2x(dx/dt)+2y(dy/dt)
y=xtan(60o)
dx/dt=√3 m/s
Solving, i get

(dy/dt)=-3 m/s
But that's not correct, the answer says its 2 m/s. I don't seem to get where i am wrong.
Any help is appreciated. :smile:

Thanks!
 
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Hi Pranav! :smile:

Pranav-Arora said:
y=xtan(60o)
dx/dt=√3 m/s
Solving, i get

(dy/dt)=-3 m/s

Are you sure about that step?

Recheck your solving,

0 = 2x\sqrt{3} + 2\sqrt{3} x \frac{dy}{dt}
 
Last edited by a moderator:
Infinitum said:
Hi Pranav! :smile:

Are you sure about that step?

Recheck your solving,

0 = 2x\sqrt{3} + 2\sqrt 3 x \frac{dy}{dt}

Oh yes, sorry, my mistake, i wrote tan60=1/√3. :-p

But now i get (dy/dt)=-1.
 
Pranav-Arora said:
Oh yes, sorry, my mistake, i wrote tan60=1/√3. :-p

But now i get (dy/dt)=-1.

Yes. That's correct. So, what's the velocity of the end B of the rod now?
 
Infinitum said:
Yes. That's correct. So, what's the velocity of the end B of the rod now?

Isn't (dy/dt) the velocity of the end B?
 
Pranav-Arora said:
Isn't (dy/dt) the velocity of the end B?

That is the vertical component of B's velocity :wink:
 
Infinitum said:
That is the vertical component of B's velocity :wink:

Ah i get it, thanks for the help Infinitum!
I shouldn't be studying Physics at midnight.
 
Pranav-Arora said:
Ah i get it, thanks for the help Infinitum!
:smile:
I shouldn't be studying Physics at midnight.

Good night, then :wink:
 
Good night, then :wink:
Hehe, no, i am posting one more question and then i will go to sleep. :D
 
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