# Homework Help: Expression for the instantaneous angular velocity

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1. May 16, 2017

### Rahulrj

1. The problem statement, all variables and given/known data
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?
2. Relevant equations
$\omega$ = $\omega_0$+$\alpha t$
$\omega^2$ = $\omega_0^2$ + 2$\alpha \theta$
$\theta$ = $\omega_0 t$+1/2$\alpha t^2$
3. 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.

Last edited: May 16, 2017
2. May 16, 2017

### ehild

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.

3. May 16, 2017

### Rahulrj

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$\omega = \frac{\omega}{\sqrt{1-\omega^2t^2}}$

4. May 16, 2017

### ehild

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

5. May 16, 2017

### Rahulrj

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

Last edited: May 16, 2017
6. May 16, 2017

### haruspex

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

7. May 16, 2017

### ehild

You have $\theta = \tan^{-1}({vt/h})$
Differentiate it with respect to t. What do you get? Remember v and h are constants.

8. May 16, 2017

### Rahulrj

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 $\omega = \frac{\omega_0}{1+\omega_0^2t^2}$

Last edited: May 16, 2017
9. May 16, 2017

### ehild

correct so far...
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?

10. May 16, 2017

### Rahulrj

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

11. May 16, 2017

### ehild

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