Rotational kinematics of circular saw

  • Thread starter MozAngeles
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  • #1
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Homework Statement


When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4740 rpm to zero in 3.00 s.
A. What is the angular acceleration of the blade(rev/s^2)? GOT IT
B. What is the distance traveled by a point on the rim of the blade during the deceleration(ft)? GOT IT
C. What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration(in)? NEED HELP

Homework Equations


[tex]\theta[/tex]=s/r
all of the angular kinematics equations

The Attempt at a Solution

i solved part be with s=[tex]\theta[/tex]*r after i had found theta using the kinematics equation ([tex]\omega[/tex]^2=[tex]\omega[/tex]0^2+2[tex]\alpha[/tex]*[tex]\Delta[/tex][tex]\theta[/tex], solving for theta using the acceleration from part A. IF you think part C is just the answer to part B in inches thats wrong, but I do not know what else to do. PLease help.
 

Answers and Replies

  • #2
Doc Al
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Imagine a circle. The top of the circle represents the starting point. Where is the end point of its motion? Draw an arrow between those points. That's the displacement.
 
  • #3
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so the net displacement is the diameter of the wheel, right? So then to i use the S- theta*r, but for r use 10?
 
  • #4
Doc Al
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so the net displacement is the diameter of the wheel, right?
Only if the net angular displacement is an odd multiple of pi radians. In other words, only if the point of the blade ends up on the opposite side of the circle after it stops.

So then to i use the S- theta*r, but for r use 10?
Not sure what you mean here.

Think about this: How many revolutions does the blade make? If it makes one complete revolution, the net displacement is zero. It's back where it started. If it makes 2.5 revolutions, it is 180 degrees away, thus the displacement equals the diameter. And so on.

Make sense?
 

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