Finding Velocity of a Rotating Light Source Along a Wall

[Abhishekdas]

Angular velocity question...

Homework Statement

Assume that a light source always emits a single thin ray of light...Now such a light source is kept in front of a long wall. The light source rotates at a constant anticlockwise angular velocity w. Now at a certain instant the light ray is r metres long and makes an angle of 45 degree with the wall(please refer diagram in attachment).
Find the velocity of the end point of the ray at this instant(along the wall).

Homework Equations

v=rw

The Attempt at a Solution

Angualar velocity * r = tangential velocity so v = rw. I found tangential velocity and i took component of this velocity along the wall so my answer is rwcos45 but answer in the book is rw sec45...I did not understand how...So please help...

Thank you in advance...

 

[Doc Al]

Start by writing an expression for the height of the endpoint as a function of angle. Hint: You'll use calculus to find the speed of the endpoint.

 

[Doc Al]

Angualar velocity * r = tangential velocity so v = rw. I found tangential velocity and i took component of this velocity along the wall so my answer is rwcos45 but answer in the book is rw sec45...

FYI, your error is treating the ray of light as if it were a swinging stick of fixed length r. It's not.

 

[Abhishekdas]

Start by writing an expression for the height of the endpoint as a function of angle. Hint: You'll use calculus to find the speed of the endpoint.

Hi Doc Al...

I did not get what you wrote by height of the endpoint as a function of angle...

And replying to your second post i am not getting the difference between a swinging stick of fixed length and the ray of light...I know the light ray can change its length but for that INSTANT i can view this problem as a fixed rod of length r rotating with angular velocity w and it hits the surface of water(analogous to the wall) and we are asked to find out the velocity of the point on the rod which is just above the surface of water or just in contact with the water surface...Please tell me what is wrong in thinking this way...

 

[Doc Al]

I know the light ray can change its length but for that INSTANT i can view this problem as a fixed rod of length r rotating with angular velocity w

Actually, you can't. The length is changing as well as the angle.

In any case, you want the speed of the spot of light as it moves along the wall. So find an expression for the height of that spot (using simple trig relationships), then you can find the rate at which that height changes.

 

[Abhishekdas]

Ya maybe both length and angle changing is causing the problem but still i am of the impression that for that instant my assumption is valid...maybe you can tell better...

And you i did what you said...i am getting the same result...

height(h) = rsin(theta)
dh/dt=rcos(theta)*d(theta)/dt

now d(theta)/dt = w so its rw cos theta

 

[Doc Al]

Ya maybe both length and angle changing is causing the problem but still i am of the impression that for that instant my assumption is valid...maybe you can tell better...

Since you are concerned with how things change, you must be careful about assumptions made "for that instant".

And you i did what you said...i am getting the same result...

height(h) = rsin(theta)

Since r is not constant, find a different expression for height h. (Hint: Use distance to the wall, which is constant.)

 

[Abhishekdas]

Ya...i did think r would not be constant but i neglected that too...my mistake anyway just check if this is correct please...

h=rcos 45 tan(theta)
dh/dt=rcos 45 *(sec theta)^2*d(theta)/dt now dtheta/dt = w.

Substituting values i get the correct answer...

I hope this is correct...

Anyway Doc Al maybe you can help me out with this... i am not saying that my calculus is weak...I am able to answer good questions based on integration and differentiation but i do get confused with what to neglect and what approxiamtions to make at times which you are telling me to be careful about...Its not funny but i do make mistakes taking "at the instant" approximations...
Like i still don't exactly know why this case was different from a rod going into a water thing...and then i thought that r is so big so i can take r constant for that instant(again) and whoa...it made a difference of cos theta becoming sec...So i don't know can you help me out with these...
I mean atleast can you give me a batter explanation of why my above approximations were wrong? I am only hoping that things will get better with practice...

Anyway thank a lot...

 

[Doc Al]

Ya...i did think r would not be constant but i neglected that too...my mistake anyway just check if this is correct please...

h=rcos 45 tan(theta)
dh/dt=rcos 45 *(sec theta)^2*d(theta)/dt now dtheta/dt = w.

Substituting values i get the correct answer...

I hope this is correct...

What allows this to work is that, despite r and θ being functions of time, the combination rcosθ is a constant. It's the distance to the wall. (This is what I was hinting at earlier.)

I'd do this:
h = D tanθ (where D = rcosθ is the constant distance to the wall)
dh/dt = D sec²θ dθ/dt
dh/dt = rcosθ sec²θ dθ/dt
dh/dt = r secθ ω

Note that there are no approximations involved here.

General advice on approximations: In many (but not all) cases the best way to use an approximation is to write the exact expression, even if it's too difficult to evaluate. Then you judiciously apply standard approximations to simplify the exact expression into something easier to work with.

(By 'standard approximations' I mean things like sinθ ≈ θ, where θ is small enough.)

 

[Abhishekdas]

What allows this to work is that, despite r and θ being functions of time, the combination rcosθ is a constant. It's the distance to the wall. (This is what I was hinting at earlier.)

I'd do this:
h = D tanθ (where D = rcosθ is the constant distance to the wall)
dh/dt = D sec²θ dθ/dt
dh/dt = rcosθ sec²θ dθ/dt
dh/dt = r secθ ω

I did the same thing...as i posted(didnt show all the working though)...So that's correct...good...

And you ...i will take your advice on approximations...i will stick to the exact thing...Because the approximations which I assume to be small turn out to be really decieving...hmmm...thanks a lot...