1. Sep 15, 2009

darko21

1. The problem statement, all variables and given/known data
A crankshaft with a diameter of 3.0 cm , rotating at 2300 rpm comes to a halt in 1.30 s . What is the tangential acceleration of a point on the surface of the crankshaft?

2. Relevant equations
W=v/r
Wf=Wi+@t
@=At/r

3. The attempt at a solution
-2300 rpm=138000r/s=13006.19m/s

-v=13006.19m/s
-r=.015m
-t=1.3

-W=867079

-@=666983

-@10004.7

Last edited: Sep 15, 2009
2. Sep 15, 2009

sylas

What are the units on those numbers?

3. Sep 15, 2009

darko21

sorry, edited

4. Sep 15, 2009

drizzle

did you convert the units, cm to m and rpm to rps?

edit: you did, but:

2300 rpm=2300r/m=2300r/(60 sec)=2300/60 r/s

Last edited: Sep 15, 2009
5. Sep 15, 2009

darko21

yes, if you see i said the radius = 1.5cm=.015m and 2300rpm=13006.19m/s

6. Sep 15, 2009

sylas

Giving names the variables so we know what is being discussed.

W (angular velocity in radians per second)
v (tangent velocity in meters per second)
@ (angular acceleration in radians per sec per sec)
A (tangential acceleration in meters per sec per sec)
t (time in seconds)

Wf is final and Wi is initial; but since it is going to zero, we can just use W = @t

Drizzle is right. You've done the unit conversions in step 3 incorrectly.

7. Sep 15, 2009

darko21

ahh finally!! im clumsy.. thank you

8. Sep 15, 2009

drizzle

no you’re not, just think logically next time after you did the conversion of units, i.e. could the number of rounds in a second be greater than the number of rounds in min of the same system or should it be less? just a bit of advice