# Speed of a projection of a rotating object (1 Viewer)

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#### caffeine

##### Guest
Someone asked me a basic physics question, and I'm not believing my answer, even though it appears to have the correct limiting behavior. This is driving me completely insane.

Suppose I stand on a sidewalk, 10 feet from the middle of an infinitely long and straight road. I have a flashlight in my hand, and I rotate the flashlight at some constant angular speed $$\omega$$. The question posed to me was "what is the speed of the light's image along the middle of the road, 30 feet from where I'm standing"?

Code:
A                B            road
* ------------- * ---------------
.
.
*O
I'm standing at point O with my flashlight. I place an origin on myself with y pointing "up" and x pointing to the right. So $$R=OA=10$$ft and $$R'=OB=30$$ft. The velocity of the lightbeam, as a function of position should be:

$$\vec{v}(\vec{r}) = \omega\,r \left[ \hat{x}\,\cos(\theta) - \hat{y}\, \sin(\theta) \right]$$

Let $$\theta$$ be measured from the vertical. At point B,

$$\theta = \cos^{-1}\left(\frac{R}{R'}\right)$$

And I'm already in trouble. I want the speed of the flashlight along the middle of the road, so I'm asked for $$\vec{v}(B)\cdot\hat{x}$$. If you plug in my $$\theta$$ into my expression for the flashlight's velocity and take the dot product, you get something constant:

$$\vec{v}(B)\cdot\hat{x} = \omega \, R$$

However, I refuse to believe this. This is saying the flashlight's x component of velocity is constant. My physical intuition says this is nonsense. I can certainly integrate the time it takes for the flashlight to go the entire (infinite) distance of the road. The time *should* be the time it take for me to rotate through $$\pi$$ radians. That time should be $$\pi / \omega$$, however, if the x component of velocity is constant, that time will be infinite.

The full answer (what is the velocity of the flashlight anywhere on the middle of the road) is:

$$\vec{v}(R') = \omega \left[\hat{x}\,R - \hat{y} \sqrt{R'^2 - R^2} \right]$$

which of course becomes $$-\infty$$ when $$R'\rightarrow\infty$$ and $$\infty$$ when $$R'\rightarrow-\infty$$. That's expected. The speed of this result is $$|\vec{v}(R')| = \omega \, R'$$, which is correct also. But how is it possible that the x component remains the same whether $$R'\rightarrow\infty$$ or $$R'\rightarrow R$$. Does that make sense?

I can't see where I went wrong. What's going on here?

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#### Galileo

Homework Helper
Aren't you just asking the velocity of an object performing uniform circular motion at a distance R' from the circle?

You can write the position as:
$$\vec x(\theta)=R'(\sin(\theta)\hat x+\cos(\theta)\hat y)$$
With $\theta(t)=\omega$.
So the velocity is:
$$\vec v(t)=\frac{d \vec x}{d\theta}\frac{d\theta}{dt}=\omega R'(\cos(\theta)\hat x-\sin(\theta)\hat y)$$
You just want to know this velocity when $R'\cos \theta = R$ and $R'\sin \theta = |AB|=\sqrt{R'^2-R^2}$ plug it in and out pops the answer. (I don't know why you projected your velocity onto $\hat x$.)

It's not treu that the x-component remains the same as R' varies, which is indeed intuitively clear. As you increase R', you also increase R. The two are not independent as is clear from how you they arise in the problem. They are related via: $R=\sqrt{R'^2-|AB|^2}$, so when R' increases, so does R.

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#### caffeine

##### Guest
Galileo said:
Aren't you just asking the velocity of an object performing uniform circular motion at a distance R' from the circle?
Not quite. The question asked: "How fast is the beam's image on the middle of the street moving when it is 30 feet from me". In other words, as I sweep my arm, there's an intersection of the lightbeam with the middle of the road. How fast is that white dot moving.

Since the road is parallel to xhat, the velocity I'm after is the component in the xhat direction.

Galileo said:
You can write the position as:
$$\vec x(\theta)=R'(\sin(\theta)\hat x+\cos(\theta)\hat y)$$
With $\theta(t)=\omega$.
So the velocity is:
$$\vec v(t)=\frac{d \vec x}{d\theta}\frac{d\theta}{dt}=\omega R'(\cos(\theta)\hat x-\sin(\theta)\hat y)$$
You just want to know this velocity when $R'\cos \theta = R$ and $R'\sin \theta = |AB|=\sqrt{R'^2-R^2}$ plug it in and out pops the answer. (I don't know why you projected your velocity onto $\hat x$.)
If you plug your cos and sin terms into your velocity expression, your velocity expression is identical to mine!

I don't think I did a good job of explaining the problem, so I'll try again. Suppose you're standing at point O, and you swing a flashlight at constant angular speed from your left to your right.
In front of you, there's an infinitely long and straight road parallel to the x axis. As you swing the flashlight, a bright dot appears on the road. At first, the dot appears very far away to the left, and as you swing your arm, the dot travels along the road (parallel to the x axis), passes through the point (0,R) right in front of you, and then moves very far down the road to the right. During the whole time, the bright dot remains on the road, and so its velocity is always in the x direction (which explains why I projected onto the x axis).

Your approach and my approach are completely equivalent. I started out by stating the expression for velocity in uniform circular motion. You differentiated the position of an object in UCM. We both have equivalent expression for sine and cosine (you explicitly stated yours, I kept mine in terms of acos, but it is equivalent). We ended up with equivalent expressions.

The difference is that I projected the velocity onto the x axis, but that's because the white dot's velocity vector _is_ in the x direction, because the white dot travels on the road.

Hope I did a better job explaining this. I really feel sheepish for asking about this problem! :(

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C

#### caffeine

##### Guest
Let $\vec{R}'$ be the position of the white dot---the intersection of the light beam with the road. Breaking $\vec{R}'$ down into components (I'm defining $\theta$ to be measured from the positive x axis up to point B):

$$R'_y = R \qquad\quad \hbox{ and } \qquad\quad R'_x = R\cot(\theta)$$

So $|\vec{R}'| = R\, \csc(\theta)$, and we can write $R'$ in vector form as:

$$\vec{R}' \;=\; R \csc(\theta) \left[ \; \hat{x}\,\cos(\theta) \;+\; \hat{y}\,\sin(\theta) \; \right] \;=\; R \; \left[ \hat{x} \cot(\theta) \; + \; \hat{y} \right]$$

Since $\vec{R}'$ describes the position of the beam on the street, its derivative will be the velocity of the beam.

$$\vec{v} \; = \; \frac{d\vec{R}'}{dt} \; = \; \frac{d\vec{R}'}{d\theta} \frac{d\theta}{dt} \; = \; -\omega R \csc^2(\theta) \, \hat{x}$$

However, $csc(\theta) = R'/R$, so:

$$\vec{v} = -\omega \frac{R'^2}{R} \, \hat{x}$$

The answer can be checked by integration. From intuition, I swing my arm from one side of my body to the other in a time $\Delta t = \pi / \omega$. The answer reproduces this result:

$$|\vec{v}| = \frac{dx}{dt} = \frac{dx}{d\theta}\frac{d\theta}{dt}$$

So:

$$dt = \frac{1}{v(\theta)} \frac{dx}{d\theta} \, d\theta$$

Doing the integration,

$$\Delta t = \int_0^{\pi} \left( \frac{1}{\omega R \csc^2(\theta)} \right) \frac{d}{d\theta} \left( R\, \cot(\theta) \right)\, d\theta = \frac{1}{\omega}\int_0^{\pi} d\theta = \frac{\pi}{\omega}$$

As expected. Also, you know the the dot will be hauling butt when $\theta=0$ and $\theta=\pi$. It turns out that this is the very behavior of $\csc^2(\theta)$.

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#### Physics Monkey

Homework Helper
Your expression for the velocity is wrong because it is orthogonal to the position, and the dot certainly isn't moving in a circle. If I may say, I think the trouble is that you forgot to differentiate $$R'$$ with respect to time. The position is $$\vec{x}(t) = R' (\sin{(\omega t)} \,\hat{x} + \cos{(\omega t)}\,\hat{y} )$$ but $$R'$$ is itself a function of time given by
$$R' = \frac{R}{\cos{(\omega t)}}$$.

This yields the following expression for position
$$\vec{x}(t) = R (\tan{(\omega t)}\, \hat{x} + \hat{y} ),$$
which correctly indicates that the dot isn't moving in the y direction. The velocity is therefore
$$\vec{v}(t) = R \omega \sec^2{(\omega t)}\, \hat{x},$$
which is, at least intuitively, the correct result.

Note also that the velocity is always positive and large in the distant past and the distant future. The total time is given by
$$\int \frac{dx}{v_x} = \int^{\pi/2}_{-\pi/2} \frac{R \sec^2{\theta} d\theta}{R \omega \sec^2{\theta}} = \frac{\pi}{\omega},$$
just as you would expect.

Also, I won't even mention relativity here (except that I just did).

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