Particle in non-inertial reference frame

[Tommy_512]

Homework Statement

particle moves in a xy plane, force F is actiong on it, proove that there exists a frame rotating with angular velozity w.z in which the equations of motion of this particle will be
x''=2wy'
y''=2wx'

Homework Equations

m-mass
xy-plane in which object moves
F=-kr
k=constant
r=position vector
x'=time derivative of x
x''=second time derivative
y'=time derivative of y
y''=second time derivative
z=unit vector

The Attempt at a Solution

it should be something simple, but i have no idea where to start

 

[Jhero]

Thank you for your post. I can help you prove that there exists a frame rotating with angular velocity w.z in which the equations of motion of the particle will be x''=2wy' and y''=2wx'.

First, let's define the variables in the problem. We have a particle moving in the xy-plane, with a force F acting on it. We also have a mass m, and a constant k, which is related to the force F. The position of the particle can be represented by the position vector r, and the time derivatives of x and y can be represented by x' and y', respectively. The second time derivatives of x and y are represented by x'' and y'', respectively. Finally, we have the unit vector z, which is in the direction of the angular velocity.

To prove that there exists a frame rotating with angular velocity w.z, we need to show that the equations of motion of the particle can be written in the form x''=2wy' and y''=2wx'. To do this, we will use the equations of motion for a particle in an xy-plane, which are given by:

x''=F/m * cos(θ)
y''=F/m * sin(θ)

Where θ is the angle between the position vector r and the x-axis. Now, we need to find a way to express these equations in terms of the angular velocity w.z. We can do this by using the relationship between the angular velocity and the position vector:

w=r x z

Where r is the magnitude of the position vector and z is the unit vector in the direction of the angular velocity. Substituting this into our equations of motion, we get:

x''=F/m * cos(θ) * (r x z)
y''=F/m * sin(θ) * (r x z)

Now, we can use the properties of the cross product to rewrite these equations as:

x''=F/m * (r x cos(θ) * z)
y''=F/m * (r x sin(θ) * z)

We can see that the terms (r x cos(θ)) and (r x sin(θ)) are equivalent to the components of the position vector in the x and y directions, respectively. Therefore, we can rewrite the equations as:

x''=F/m * (rx * z)
y''