What is the tangential acceleration of a rotating crankshaft?

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Homework Help Overview

The problem involves a crankshaft with a specified diameter that is rotating at a given speed and comes to a halt over a set duration. The objective is to determine the tangential acceleration of a point on the surface of the crankshaft.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss unit conversions, specifically from centimeters to meters and revolutions per minute to revolutions per second. There are attempts to clarify the relationships between angular velocity, tangential velocity, and acceleration.

Discussion Status

There is ongoing clarification regarding the correct unit conversions and the definitions of variables involved in the calculations. Some participants have provided guidance on how to approach the problem, but no consensus has been reached on the solution.

Contextual Notes

Participants have noted potential errors in unit conversions and the importance of logical reasoning in the context of the problem. The original poster's calculations have been questioned, particularly regarding the derived values for angular velocity and tangential velocity.

darko21
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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.
 
Last edited:
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darko21 said:
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?
 
sylas said:
What are the units on those numbers?

sorry, edited
 
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
 
Last edited:
yes, if you see i said the radius = 1.5cm=.015m and 2300rpm=13006.19m/s
 
darko21 said:

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.
 
ahh finally! I am clumsy.. thank you
 
darko21 said:
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 :wink:
 

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