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