# Centrifugal force

1. Apr 26, 2012

### azizlwl

"Centrifugal force"

1. The problem statement, all variables and given/known data

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 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
$Multiplying\ by\ \dot r dt=dr \ \ and\ integrating$

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

2. Relevant equations

3. The attempt at a solution
Blurr at differential equation.

2. Apr 26, 2012

### ehild

Re: "Centrifugal force"

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

3. Apr 26, 2012

### azizlwl

Re: "Centrifugal force"

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

4. Apr 26, 2012

### ehild

Re: "Centrifugal force"

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

ehild

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