Finding the acceleration of a variable crank mechanism.

[Nickel2115]

I've attempted multiple times, but don't really know which way to go and none of them are getting to the answer. I've got the question (below), got it wrong and been told the correct answer.

A variable crank mechanism consists of a slider in a slot which is rotated around an axis at constant speed 120 rev/min. The slider is moving outwards relative to the slot at constant velocity 9 m/s. What is the magnitude of its acceleration (in m/s^2) at the instant when its radius is 0.3 m?


tangential velocity v = ωr

maybe the equations of motion

v^2=u^2 + 2 x a x s



I know i have to convert the speed 120rev/min into rad/s by (120*2∏)/60 = 4∏.
Then a=W^2 x r.
so a = 16 x ∏^2 x 0.3=47.374...

The answer I'm supposed to be getting is 231.1 m/s^2

 

[TSny]

Welcome to PF!

You've calculated the centripetal component of the acceleration. There is also a tangential component that you will need to find.

Polar coordinates are appropriate for this problem. There are well-known expressions for the components of acceleration in polar coordinates. Are you familiar with them?

 

[Nickel2115]

Thanks.
I realized there was a tangential acceleration but didn't know the equations for the components of acceleration in polar coordinates.

 

[TSny]

A derivation of acceleration in polar coordinates is here.

 

[Nickel2115]

ok I've solver it now thank you