Particle Motion in a Vertical Plane: Trajectory Equation & Curve

[lfdahl]

A particle moves in a vertical plane from rest under the influence of gravity
and a force perpendicular to and proportional to its velocity. Obtain the equation of the
trajectory, and identify the curve.

 

[lfdahl]

Hint:

Spoiler

The equations of motion are:

\[m\ddot{y} = mg-k\dot{x},\: \: \: m\ddot{x} = k\dot{y}\]

One dot means a differentiation with respect to t (time): $\frac{dy}{dt} = \dot{y}.$

 

[lfdahl]

Suggested solution:

Spoiler

Assume the mass is dropped from $x=y=0$ at $t=0$. Let the $y$-axis be positive down. We then have:
(Force = k velocity and physics notation: $\dot{y} = \frac{dy}{dt}$)

\[m\ddot{y} = mg-k\dot{x},\: \: \: m\ddot{x} = k\dot{y}\]

Or
\[\ddot{y} = g-w\dot{x},\: \: \: \ddot{x} = w\dot{y},\: \: \: w = k/m.\]

Substituting gives: \[\frac{\mathrm{d}^2\dot{y} }{\mathrm{d} t^2} =-w^2\dot{y}\]

Or

\[\dot{y} = B \sin (wt)\Rightarrow y = -\frac{B}{w}\cos (wt)+\frac{B}{w} \\\\ \dot{x}=-B\cos (wt)+B \Rightarrow x = -\frac{B}{w} \sin (wt)+Bt\]
Substitution in original differential equation gives: $B = g/w$.

Finally

\[x = -\frac{g}{w^2} \sin (wt)+\frac{g}{w}t \\\\ y = -\frac{g}{w^2}\cos (wt)+\frac{g}{w^2}\]

These determine a cycloid produced by a wheel of radius $R = g/w^2$ rolling on the $x$-axis at speed $\frac{g}{w}$, where $w = k/m$.

 

[solakis1]

A particle moves in a vertical plane from rest under the influence of gravity
and a force perpendicular to and proportional to its velocity. Obtain the equation of the
trajectory, and identify the curve.

Spoiler

Since the particle moves from rest shouldn't we have : at t=0 ,dy/dt=dx/dt=x=y=0 ??

 

[lfdahl]

Since the particle moves from rest shouldn't we have : at t=0 ,dy/dt=dx/dt=x=y=0 ??

Spoiler

Yes, and as far as I can see, these conditions are fulfilled:

\[\dot{x}(0) = -\frac{g}{w}\cos (0)+\frac{g}{w} = 0 \\\\ \dot{y} = \frac{g}{w}\sin (0) =0\]

Likewise:

$x(0) = y(0) = 0$.

 

[solakis1]

Hint:

Spoiler

The equations of motion are:

\[m\ddot{y} = mg-k\dot{x},\: \: \: m\ddot{x} = k\dot{y}\]

One dot means a differentiation with respect to t (time): $\frac{dy}{dt} = \dot{y}.$

I would like to know how you got the equations of motion ,if it is possible

Thanks

 

[lfdahl]

I would like to know how you got the equations of motion ,if it is possible

Thanks

Spoiler

The two only forces involved in the cycloid motion are the gravitational force, and a velocity dependent force.
The former is in our coordinate system simply:

\[\vec{F_g} = m\binom{0}{g}\]

The latter force is proportional to the particle velocity, $\vec{v}= \binom{\dot{x}}{\dot{y}}$, and perpendicular to it. Hence we have:

\[\boldsymbol{F}_v = -k \hat{\mathbf{v}} = -k\binom{-\dot{y}}{\dot{x}} = k\binom{\dot{y}}{-\dot{x}}\]

Using Newtons 2nd law:

\[\boldsymbol{F}_{res} = m\binom{\ddot{x}}{\ddot{y}} = \boldsymbol{F}_g + \boldsymbol{F}_v = m\binom{0}{g} +k\binom{\dot{y}}{-\dot{x}}\]

In component form:

\[m\ddot{x} = k\dot{y}\\\\ m\ddot{y} = mg - k \dot{x}\]

 

[solakis1]

Spoiler

The two only forces involved in the cycloid motion are the gravitational force, and a velocity dependent force.
The former is in our coordinate system simply:

\[\vec{F_g} = m\binom{0}{g}\]

The latter force is proportional to the particle velocity, $\vec{v}= \binom{\dot{x}}{\dot{y}}$, and perpendicular to it. Hence we have:

\[\boldsymbol{F}_v = -k \hat{\mathbf{v}} = -k\binom{-\dot{y}}{\dot{x}} = k\binom{\dot{y}}{-\dot{x}}\]

Using Newtons 2nd law:

\[\boldsymbol{F}_{res} = m\binom{\ddot{x}}{\ddot{y}} = \boldsymbol{F}_g + \boldsymbol{F}_v = m\binom{0}{g} +k\binom{\dot{y}}{-\dot{x}}\]

In component form:

\[m\ddot{x} = k\dot{y}\\\\ m\ddot{y} = mg - k \dot{x}\]

Spoiler

Isn't the vector: (\(\displaystyle -k\dot y,k\dot x\)),also perpenticular to the particle velocity and proportional to it ??

 

[Greg]

Please post anything relating to the solution of a problem (in the Challenges forum) in [sp]...[/sp] tags. Thanks. :)

 

[lfdahl]

Spoiler

Isn't the vector: (\(\displaystyle -k\dot y,k\dot x\)),also perpenticular to the particle velocity and proportional to it ??

Spoiler

Yes, this vector also fulfills the condition, and you can elaborate on the equations of motion with opposite signs for the velocity dependent force. You´ll see, that the y-component turns out to be the same as before. Only the x-component changes sign, i.e. the direction of motion would be from right to left starting in origo.

 

[solakis1]

Spoiler

Yes, this vector also fulfills the condition, and you can elaborate on the equations of motion with opposite signs for the velocity dependent force. You´ll see, that the y-component turns out to be the same as before. Only the x-component changes sign, i.e. the direction of motion would be from right to left starting in origo.

Spoiler

Thanks
I have done this allready