Polar Coordinates [Finding the velocity]

AI Thread Summary
The discussion focuses on calculating the speed of a projectile using polar coordinates, given specific radar readings. The initial values include an angle of 30 degrees, a radial distance of 2000 meters, and rates of change for distance and acceleration. Participants explore the equations for radial and angular velocities, but there is confusion regarding the correct application of these equations. The conversation emphasizes the need to differentiate the relationship between Cartesian and polar coordinates and correctly apply the second derivative to find the projectile's speed, ultimately leading to the confirmed answer of 299.7 m/s.
MrMechanic
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Homework Statement


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]


Homework Equations


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

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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MrMechanic said:

Homework Statement


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]

Homework Equations


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

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.

Could you show please, your work in detail?
And what do you mean on V^2 = Vr + Vθ? The square of a vector is not the sum of its components.

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
 
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
 
what are the values of x and its derivatives... same goes for y.. I got confused. But thanks sir.
 
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
 
Follow the advice from ehild (somehow I had the impression ehild is a she).
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:
cos30=x/2000 --> x = 1732.05
OK
dR/dt = 1732.05 (dθ/dt)(secθtanθ)
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.
 
How to find the acceleration?
 
@Awreal
MrMechanic said:
what are the values of x and its derivatives... same goes for y.. I got confused. But thanks sir.
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##.
 
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