# Polar Coordinates [Finding the velocity]

1. Sep 6, 2014

### MrMechanic

1. The problem statement, all variables and given/known data
The projectile A is being tracked by the radar at O. At a given instant,
the radar readings are θ = 30degrees, R = 2000m, dR/dt = 200 m/s, and d^2R/dt^2 = 20 m/s^2.
Determine the speed of the projectile at that instant.
THE ANSWER AT THE BACK IS 299.7m/s
[PLEASE SEE ATTACHMENT FOR PROBLEM & FIGURE]

2. Relevant equations
I've used the Vr = dR/dt & Vθ= R(dθ/dt)

3. The attempt at a solution
I've worked out an equation which is
cos30=x/2000 --> x = 1732.05
dR/dt = 1732.05 (dθ/dt)(secθtanθ)
200 = 1732.05 (dθ/dt)(sec30tan30)
dθ/dt = 0.1732 deg/sec?
And using V^2 = Vr + Vθ I didn't get 299.7m/s So I'am wrong.

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2. Sep 6, 2014

### ehild

And what do you mean on V^2 = Vr + Vθ???? The square of a vector is not the sum of its components.

ehild

3. Sep 7, 2014

### ehild

The problem is very interesting, I want to keep it alive.

R(t) is the time-dependent distance of the projectile from the origin. With the Cartesian coordinates of the projectile, $R^2=x^2+y^2$. Differentiate the equation twice.

ehild

4. Sep 7, 2014

### ehild

Differentiating the equation $R^2=x^2+y^2$ results $R\dot R=x\dot x+y\dot y$. The second derivative is ${\dot R }^2 +R\ddot R={\dot x}^2+{\dot y}^2+ x\ddot x +y\ddot y$.
$R$, $\dot R$ and $\ddot R$ are given. It is a projectile, the horizontal acceleration is zero, the vertical one is -g. And ${\dot x}^2+{\dot y}^2=v^2$, the square of the speed.

ehild

5. Sep 8, 2014

### MrMechanic

what are the values of x and its derivatives... same goes for y.. I got confused. But thanks sir.

6. Sep 8, 2014

### ehild

The x,y coordinates of the projectile can be expressed with the polar coordinates R and θ, how? You gave already the x coordinate as x=Rcos(30°), what is y?
The motion of a projectile is the resultant of a horizontal motion and with a vertical one. The only force is the gravity of Earth. What are the horizontal and vertical accelerations?

ehild

7. Sep 8, 2014

### BvU

You get the solution handed to you on a silver platter: in the expression for the second derivative, you know everything except v:
you don't need x, because it is multiplied with its second derivative, which is zero
you don't need $\dot x$ or $\dot y$: the sum of squares is v square and v is what you are after!
all you need is to calculate y, which is almost trivial!

Commenting on the equations in your solution:
OK
Looks like $R \tan\theta \; \dot\theta$ and I don't follow that. So from there on, you're lost as far as I can see.

@ehild: sorry to have jumped in; thought MrM was awake and you were away; it's the other way around now.
Nice exercise indeed.

8. Nov 29, 2016

### Awreal

How to find the acceleration?

9. Nov 30, 2016

### ehild

@Awreal
You should know how the Cartesian coordinates and the polar ones are related. x=Rcos(θ), y=Rsin(θ).
You also know that the horizontal component of acceleration of the projectile is zero, $\ddot x =0$, the vertical component is $\ddot y = - g$. Use these values, together with the given data in the expression of the second derivative of R2, ${\dot R }^2 +R\ddot R={\dot x}^2+{\dot y}^2+ x\ddot x +y\ddot y$.