What is the tangential acceleration of a rotating crankshaft?

[darko21]

Homework Statement

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?

Homework Equations

W=v/r
Wf=Wi+@t
@=At/r

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

the answer is completely wrong.

 

[sylas]

A crankshaft with a diameter of 3.0 , rotating at 2300 comes to a halt in 1.30 . What is the tangential acceleration of a point on the surface of the crankshaft?

What are the units on those numbers?

 

[darko21]

What are the units on those numbers?

sorry, edited

 

[drizzle]

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

edit: you did, but:

-2300 rpm=138000r/s=...?

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

 

[darko21]

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

 

[sylas]

Homework Statement

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?

Homework Equations

W=v/r
Wf=Wi+@t
@=At/r

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)
r (radius in meters)
@ (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.

 

[darko21]

ahh finally! I am clumsy.. thank you

 

[drizzle]

ahh finally! I am clumsy.. thank you

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