How to get ##\ddot r## when you have ##r##, ##\theta## and right trig

In summary: You can probably calculate it if you know the values of the acceleration and the velocity, but it's not necessary to solve the equation for the angle.
  • #1
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Homework Statement
A radar detects that a particle, located ##40 m## above the ground, moves towards it with velocity ##v_r=5 \frac{m}{s}##. The modulus of ##\vec v## is constant, and the modulus of the acceleration at that moment is ##a=10 \frac{m}{s}##. The radius vector that goes from the radar to the particle forms an angle of ##60°## with the ground. Find ##\ddot r##
Relevant Equations
##sin(x)=\frac{op}{hyp}##
I have a right triangle: one of the angles is ##60°## (that's ##\theta##), one of the sides is ##40 m## long, and the hypotenuse is equal to the radius. Now I can find an expression for ##r## and that expression is ##r=\frac{height}{sin \theta}##. If I differentiate it, I'll get ##\dot r## and if I differentiated it again I would get ##\ddot r##. Now, how can I differentiate that expression? because time doesn't appear in the expressions.
 
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  • #2
Like Tony Stark said:
Homework Statement: A radar detects that a particle, located ##40 m## above the ground, moves towards it with velocity ##v_r=5 \frac{m}{s}##. The modulus of ##\vec v## is constant, and the modulus of the acceleration at that moment is ##a=10 \frac{m}{s}##. The radius vector that goes from the radar to the particle forms an angle of ##60°## with the ground. Find ##\ddot r##
Homework Equations: ##sin(x)=\frac{op}{hyp}##

I have a right triangle: one of the angles is ##60°## (that's ##\theta##), one of the sides is ##40 m## long, and the hypotenuse is equal to the radius. Now I can find an expression for ##r## and that expression is ##r=\frac{height}{sin \theta}##. If I differentiate it, I'll get ##\dot r## and if I differentiated it again I would get ##\ddot r##. Now, how can I differentiate that expression? because time doesn't appear in the expressions.
This looks like a continuation of a problem you posted earlier. In that problem, I think you were looking for the angle between the velocity vector and a vector that is perpendicular to the hypotenuse of your right triangle. If you know that angle and the target's acceleration vector, then you can decompose the acceleration vector into two orthogonal components, one of which is the ##\ddot r## that you are looking for here. Edit: This assumes that the target's motion and the radar are all in the same plane.
 
  • #3
Is the angle of 60 deg. constant or variable?
 
  • #4
tnich said:
This looks like a continuation of a problem you posted earlier. In that problem, I think you were looking for the angle between the velocity vector and a vector that is perpendicular to the hypotenuse of your right triangle. If you know that angle and the target's acceleration vector, then you can decompose the acceleration vector into two orthogonal components, one of which is the ##\ddot r## that you are looking for here. Edit: This assumes that the target's motion and the radar are all in the same plane.

Yess, it's another part of the same problem. Because I could do the first one (the one of the angle) with your help but I got stuck in the other one. The thing is that I don't know how to get the angle formed by the acceleration and the axis
 
  • #5
Like Tony Stark said:
Yess, it's another part of the same problem. Because I could do the first one (the one of the angle) with your help but I got stuck in the other one. The thing is that I don't know how to get the angle formed by the acceleration and the axis
You know the magnitude of the velocity is constant. What does that tell you about the angle between the velocity vector and the acceleration vector?
 
H2: What is the formula for finding ##\ddot r## when given ##r##, ##\theta##, and right trig?

The formula for finding ##\ddot r## when given ##r##, ##\theta##, and right trig is: ##\ddot r = r - \dot r^2/r##.

H2: How do I convert from polar coordinates to Cartesian coordinates?

To convert from polar coordinates to Cartesian coordinates, use the following formulas: ##x = r\cos\theta## and ##y = r\sin\theta##. These formulas will give you the x and y coordinates in the Cartesian plane.

H2: What is the difference between ##\dot r## and ##\ddot r##?

##\dot r## and ##\ddot r## are both related to the velocity of an object in polar coordinates. However, ##\dot r## represents the rate of change of the distance from the origin, while ##\ddot r## represents the rate of change of the velocity in the radial direction.

H2: How do I find the acceleration in polar coordinates?

To find the acceleration in polar coordinates, you will need to use the formula: ##\ddot r = r\dot\theta^2 - r\cos\theta\dot\theta##. This formula takes into account both the radial and tangential components of acceleration.

H2: How does right trigonometry help in solving problems involving polar coordinates?

Right trigonometry is essential in solving problems involving polar coordinates because it helps to relate the polar coordinates to the Cartesian coordinates. It also allows for the conversion between the two coordinate systems, making it easier to solve complex problems involving both types of coordinates.

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