# Physics/Math question

#### Eeezee

##### Diamond Member
In cylindrical coordinates, the velocity in the theta direction is as follows

v_theta = r*(dtheta/dt)

If you take the time derivative of this, you get the acceleration in the theta direction

a_theta = (dr/dt)*(dtheta/dt)+r*(d^2 theta / dt^2)

This is wrong. Can anyone tell me why? The accepted answer is

a_theta = 2*(dr/dt)*(dtheta/dt)+r*(d^2 theta / dt^2)

Where does that factor of 2 come from?

Edit: The 2 pops up because the unit vectors in cylindrical coordinates are time-dependent too. Oops! Been too long since I've taken vector calc I guess

#### Eeezee

##### Diamond Member
The lack of responses has pwnd me!

#### TuxDave

##### Lifer
I'm still trying to figure out why there's a 2 there also.

#### JohnCU

##### Banned
hold on, i'm burned out but give me a second.

#### Eeezee

##### Diamond Member
the theta component for position is rho*theta, taking two time derivatives gives you that factor of two but now the expression is missing the (d^2 rho/dt^2) term! I'm certain that the expression is correct because it's used in kinematics and in real devices.

*scratches brain*

#### JohnCU

##### Banned
is r, the radius, not a constant, ie it doesnt change with time?

oh we are doing linear acceleration?

#### Eeezee

##### Diamond Member
r, the radial difference, definitely changes in time.

#### JohnCU

##### Banned
Originally posted by: Eeezee
r, the radial difference, definitely changes in time.

yeah im tired and suck at mechanical stuff but i'm trying

#### TheoPetro

##### Banned
try integration that helps me sometimes.

taking the time derivative would be d/dt ( r * dtheta/dt) or [d( r * dtheta/dt)]/dt or d(r * dtheta)/dt^2

its really late here tho (3am) so this may not make any sense

#### Eeezee

##### Diamond Member
I found a site that helped a lot

Text

Done

#### JohnCU

##### Banned
Originally posted by: Eeezee
I found a site that helped a lot

Text

Done

ahhhhhhhhhh that's why i'm EE instead of ME, too hard.