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

[Saitama]

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 60^o with the ground.

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(60^o)
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.

Thanks!

 

[Infinitum]

Hi Pranav!

y=xtan(60^o)
dx/dt=√3 m/s
Solving, i get

(dy/dt)=-3 m/s

Are you sure about that step?

Recheck your solving,

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

 

[Saitama]

Hi Pranav!

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.

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

 

[Infinitum]

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

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

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

 

[Saitama]

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?

 

[Infinitum]

Isn't (dy/dt) the velocity of the end B?

That is the vertical component of B's velocity

 

[Saitama]

That is the vertical component of B's velocity

Ah i get it, thanks for the help Infinitum!
I shouldn't be studying Physics at midnight.

 

[Infinitum]

Ah i get it, thanks for the help Infinitum!

I shouldn't be studying Physics at midnight.

Good night, then

 

[Saitama]

Good night, then

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