Angular deceleration

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Homework Statement



A circular saw blade completes 1200 revolutions in 40s while coasting to a stop after being turned off. Assuming constant deceleration, what are the angular deceleration, and the initial angular speed??

Homework Equations


v=wr
w^2=w0^2 + 2alpha*delta(theta)





i can't figure out how to start this problem off.
 

Answers and Replies

  • #2
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You need 2 kinematics that have time as a variable
 
  • #3
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You need 2 kinematics that have time as a variable
yeah...could you possibly elaborate for me???
 
  • #4
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What kinematic equations do you know?
 
  • #5
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What kinematic equations do you know?
all of them...and they are in my book too. But i think i solved the initial angular speed, but i don't know how to find the angular deceleration...without using time.
 
  • #6
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but i don't know how to find the angular deceleration...without using time.
Time is given, why wouldnt you want to use it???
 
  • #7
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You need to know how long it takes for it to come to a stop after turning it off.
 
  • #8
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isn't that 40s then?
 
  • #9
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Without the initial speed, you can't calculate the angular acceleration. Are you sure you don't have any other information?
 
  • #10
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which is 188rad/sec i think. I took (1800revs/min)(2pie rad/rev)(60s/min) to get 188rad/sec. Isn't that the initial angular speed/velocity???
 
  • #11
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Without the initial speed, you can't calculate the angular acceleration. Are you sure you don't have any other information?
that is the entire problem.
 
  • #12
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which is 188rad/sec i think. I took (1800revs/min)(2pie rad/rev)(60s/min) to get 188rad/sec. Isn't that the initial angular speed/velocity???
... 1800revs/min, you never mentioned that before. Is it in the problem?
 
  • #13
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no, i figured it out...i just showed you the work.
 
  • #14
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Figured it out??? The disk comes to a stop in 40 sec. This means that the velocity constantly decelerates during those 40 seconds, and 1200/40 is just the average angular velocity during those 40 seconds. This has nothing to do with the actual initial angular velocity.
 
  • #15
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Figured it out??? The disk comes to a stop in 40 sec. This means that the velocity constantly decelerates during those 40 seconds, and 1200/40 is just the average angular velocity during those 40 seconds. This has nothing to do with the actual initial angular velocity.
ok, then do you have any ideas???
 
  • #16
hage567
Homework Helper
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This is how I would approach this question:

You don't need to know the initial angular speed. What you need to do is find two kinematic equations that link the things you know and the things you don't. You want angular acceleration (alpha). You have the time it takes the blade to stop, the number of revolutions (and therefore the angular displacement) it goes before it stops, and you know the final angular speed (it's zero). You can put the initial angular speed in terms of the other variables so you don't need to know it directly. Put the two equations together, and solve for alpha.
 
  • #17
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the question asks for initial angular speed also...but i kind of know where your going with this.
 
  • #18
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i don't quite get the angular displacement...if it spends 1200 times...its going to end up in the same place it was spinning from. So technically the dispacement is 0 right??
 
  • #19
hage567
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OK, maybe think of it as angular distance. The point is, you can find out how many radians it goes through while it is decelerating to rest.

Once you find alpha, you can solve for the initial angular speed.
 
  • #20
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hage567, I believe you're wrong. Angular velocity and acceleration are the equivalent of linear acceleration and velocity along the circumference of the unit circle. This means that if we represent the angle covered by a revolving body on a straight line, we would have case of linear acceleration. For example, if we have the number of revolutions per sec before deceleration, [tex]k[/tex] we would have an angular velocity of [tex]v = 2{\pi}k[/tex]. If the body came to a stop in 40 sec, the angle covered is [tex]1400*2{\pi}[/tex]

The angular acceleration, [tex]a[/tex], would then depend on of this velocity since we have by linear acceleration,

[tex]1400*2{\pi} = \frac{1}{2}a 40^{2} + v*40[/tex]

Obviously we see that the acceleration is dependent on the choice of the initial velocity.
 
  • #21
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werg...which equation did you use to get that?? theta=theta0+w0t + 1/2at^2??
 
  • #22
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Are you familiar with this formula

[tex]d = \frac{1}{2}at^{2} + v_{i}t [/tex]

???

This is distance in function of the time for a uniformly accelerated body in straight line. What I did was to interpret the motion around the unit circle as a motion on a straight line which length is 2pi*number of revolutions. And yes, it is equivalent to the formula you just mentioned.
 
  • #23
hage567
Homework Helper
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I'm sorry Werg22, I don't understand what you're getting at. The problem states that the blade goes through 1200 rev in the 40 s. So we know how many radians it goes through during that time. It's analogous to kinematic equations for linear motion.
 
  • #24
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yes, i am familiar, i just misunderstood. Ok, so now i have to variables...a or angular acceleration and the initial angular velocity. I'm still stumped. I'm thinking of using w=w0 + alpha*t and plugging that in, but i can't seem to get it to work.
 
  • #25
hage567
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yes, i am familiar, i just misunderstood. Ok, so now i have to variables...a or angular acceleration and the initial angular velocity. I'm still stumped. I'm thinking of using w=w0 + alpha*t and plugging that in, but i can't seem to get it to work.
Are you remembering that w is zero since it comes to rest.
 

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