Centrifugal Force and the Path of a Moving Particle in a Rotating Tube

[azizlwl]

"Centrifugal force"

Homework Statement

A smooth horizontal tube of length l rotates about a vertical axis. A particle placed at the extreme end of the tube is projected towards O(axis) with a velocity lω while at the same time the tube rotates about the axis with constant angular speed ω. Determine the path.

Solution
The only force on the particle is the inertial force("Centrifugal force") mrω²

Newton's second law becomes
m\ddot r=0+mrw^2 \ or \ \ddot r=ω^2r

This part I don't understand
Multiplying\ by\ \dot r dt=dr \ \ and\ integrating

\frac {1}{2}\int d(\dot r)^2=w^2 \int rdr
\frac {1}{2}(\dot r)^2=\frac {1}{2}w^2 r^2 + c

Homework Equations

The Attempt at a Solution

Blurr at differential equation.

 

[ehild]

The (radial) velocity is the time derivative of r: \dot r= dr/dt
You can multiply the equation \ddot r= \omega^2 r by \dot r and you get \ddot r \dot r = \omega^2 r \dot r. The left hand side is the time derivative of (\dot r)^2 /2, the right hand side is the time derivative of \omega^2 r ^2 /2. So integrating both sides of the equation leads to \frac {1}{2}(\dot r)^2=\frac {1}{2}w^2 r^2 + c .

The formal procedure is that the equation \ddot r \dot r = \omega^2 r \dot r is written as \dot r d\dot r/dt = \omega^2 r dr/dt and multiplied by dt and then putting integral signs in front of both sides:
\int{\ddot r d \dot r}= \int{\omega^2 r dr}

ehild

 

[azizlwl]

] .

The formal procedure is that the equation \ddot r \dot r = \omega^2 r \dot r is written as \dot r d\dot r/dt = \omega^2 r dr/dt and multiplied by dt and then putting integral signs in front of both sides:
\int{\ddot r d \dot r}= \int{\omega^2 r dr}

ehild

The formal method you mentioned is easier for me to understand.
Thank you

 

[ehild]

OK. Remember that procedure, it is applied quite often.

ehild