Expression for the instantaneous angular velocity

[Rahulrj]

Homework Statement

The directed beam from a small but powerful searchlight placed on the ground tracks a small plane flying horizontally at a fixed height h above the ground with a uniform velocity v. If the search light starts rotating with an instantaneous angular velocity $\omega_0$ at time $t=0$ when the plane was directly overhead, then at a later time t, its instantaneous angular velocity $\omega(t)$ is given by?

Homework Equations

$\omega$ = $\omega_0$+$\alpha t$
$\omega^2$ = $\omega_0^2$ + 2$\alpha \theta$
$\theta$ = $\omega_0 t$+1/2$\alpha t^2$

The Attempt at a Solution

I am not sure where to begin the solution other than the equations I have written above.
The answer given is $\frac{\omega_0}{1+\omega_0^2t^2}$. I tried differentiating but I don't seem to get anywhere with it.

 

[ehild]

Homework Statement

The directed beam from a small but powerful searchlight placed on the ground tracks a small plane flying horizontally at a fixed height h above the ground with a uniform velocity v. If the search light starts rotating with an instantaneous angular velocity $\omega_0$ at time $t=0$ when the plane was directly overhead, then at a later time t, its instantaneous angular velocity $\omega(t)$ is given by?

Homework Equations

$\omega$ = $\omega_0$+$\alpha t$
$\omega^2$ = $\omega_0^2$ + 2$\alpha t$
$\theta$ = $\omega_0 t$+1/2$\alpha t^2$

The Attempt at a Solution

I am not sure where to begin the solution other than the equations I have written above.
The answer given is $\frac{\omega_0}{1+\omega_0^2t^2}$. I tried differentiating but I don't seem to get anywhere with it.

You tried differentiating - what?
ω need not be either a linear or quadratic function of time.
Make a picture, and everything will be clear.
How does the angle theta depend on time? How is the angular velocity ω defined?

The blue line represent the light beam at t=0, the red line is the light beam at a later time t.

 

[Rahulrj]

You tried differentiating - what?
ω need not be either a linear or quadratic function of time.
Make a picture, and everything will be clear.
How does the angle theta depend on time? How is the angular velocity ω defined?
View attachment 203608The blue line represent the light beam at t=0, the red line is the light beam at a later time t.

That to an extent was helpful however embarrassing to me I still couldn't get the right answer.
I took sine of the angle theta, so $\sin\theta = vt/l$ where l is the line connecting the searchlight and the plane
$\theta = \sin^{-1}(vt/l)$ I used $v = l\omega$ to make the substitution in vt/l
differentiating it I get$\omega = \frac{\omega}{\sqrt{1-\omega^2t^2}}$

 

[ehild]

That to an extent was helpful however embarrassing to me I still couldn't get the right answer.
I took sine of the angle theta, so $\sin\theta = vt/l$ where l is the line connecting the searchlight and the plane
$\theta = \sin^{-1}(vt/l)$ I used $v = l\omega$ to make the substitution in vt/l
differentiating it I get$\omega = \frac{\omega}{\sqrt{1-\omega^2t^2}}$

l is changing, use vt/h = tan (theta) instead.

 

[Rahulrj]

l is changing, use vt/h = tan (theta) instead.

I did use Tan at first however I got an answer with the terms l and h which couldn't figure out a way to eliminate. So with tan I get
$\theta = \tan^{-1}{vt/h}$
on differentiating I get $\omega = \frac{\omega }{h/l+l\omega^2t^2/h}$
and I do not know how to simplify it further hence I used sine in my previous post

 

[haruspex]

I did use Tan at first however I got an answer with the terms l and h which couldn't figure out a way to eliminate. So with tan I get
$\theta = \tan^{-1}{vt/h}$
on differentiating I get $\omega = \frac{\omega }{h/l+l\omega^2t^2/h}$
and I do not know how to simplify it further hence I used sine in my previous post

That doesn't make much sense. Did you mean ω₀ on the right?
Please post your working.

 

[ehild]

I did use Tan at first however I got an answer with the terms l and h which couldn't figure out a way to eliminate. So with tan I get
$\theta = \tan^{-1}{vt/h}$
on differentiating I get $\omega = \frac{\omega }{h/l+l\omega^2t^2/h}$
and I do not know how to simplify it further hence I used sine in my previous post

I don not follow you.
You have $\theta = \tan^{-1}({vt/h})$
Differentiate it with respect to t. What do you get? Remember v and h are constants.

 

[Rahulrj]

I don not follow you.
You have $\theta = \tan^{-1}({vt/h})$
Differentiate it with respect to t. What do you get? Remember v and h are constants.

Differential of $Tan^{-1}x = \frac{1}{1+x^2}$
so then $\omega = (1/1+(vt/h)^2 )* v/h$
substituting $v=l\omega$
$\omega = 1/1+(l\omega t/h)^2 * v/h$
and I end up with $\omega = \frac{\omega }{h/l+l\omega^2t^2/h}$

I found where I made mistake 'v' should be, $v=h\omega_0$
and with that substitution
I get the answer $\omega = \frac{\omega_0}{1+\omega_0^2t^2}$

 

[ehild]

Differential of $Tan^{-1}x = \frac{1}{1+x^2}$
so then $\omega = (1/1+(vt/h)^2 )* v/h$

correct so far...

substituting $v=l\omega$

That is wrong. You've just derived the expression for ω, it is not v/L.
You have to find ω₀, the angular velocity at t=0. Substitute t=0 into the equation for ω, what do you get?

 

[Rahulrj]

correct so far...

That is wrong.
You have to find ω₀, the angular velocity at t=0. Substitute t=0 into the equation for ω, what do you get?

Seems like you did not see my update in post #8.

 

[ehild]

Seems like you did not see my update in post #8.

Well, I did not see. You got it at last!
Next time remove the wrong part from your post.