Angular Deceleration Homework: 1200 Revs in 40s

[Werg22]

i did end up with 60/89 or .674 to be my angular velocity...does that make sense.?..i really don't want to write down all my steps.

No. The initial velocity is approximately 188 and 0.674 < 188. 188 is inferior to the initial velocity since it is the average during deceleration. This makes 0.674 way off.

 

[hage567]

Werg22, I will send you my solution by PM. Please let me know what you don't like about my method.

 

[Werg22]

No problem.

 

[xXmarkXx]

Werg22, I will send you my solution by PM. Please let me know what you don't like about my method.

do you have any other suggestions for me?

allright, so you posted 2400pie=1/2alpha40^2 + w₀40
and i used -w₀=alpha*40

when i plug that into solve for alpha, i keep getting stuck.

 

[xXmarkXx]

i have now arrived at 40pie revolutions/sec as the angular velocity. agree?? disagree??

 

[hage567]

i have now arrived at 40pie revolutions/sec as the angular velocity. agree?? disagree??

It is best to make sure everything is expressed in RADIANS, not revolutions, before you start. Angular speed is expressed as rad/s.

So I disagree.

 

[xXmarkXx]

It is best to make sure everything is expressed in RADIANS, not revolutions, before you start. Angular speed is expressed as rad/s.

So I disagree.

ok, i ended up at 40pie=w₀ i just labeled it incorrectly. But you still disagree?? if so, could you point me in the correct direction.

 

[Werg22]

hage567, wait up, you're solution is wrong all around, I'm writing you why.

 

[xXmarkXx]

anybody still willing to help me out?

 

[xXmarkXx]

hage, where did you get 1400*2pie...is it not 1200 revolutions? in 40 sec? so wouldn't it be 1200*2pie?

 

[hage567]

hage, where did you get 1400*2pie...is it not 1200 revolutions? in 40 sec? so wouldn't it be 1200*2pie?

I don't think I ever said that.

 

[xXmarkXx]

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, $$k$$ we would have an angular velocity of $$v = 2{\pi}k$$. If the body came to a stop in 40 sec, the angle covered is $$1400*2{\pi}$$

The angular acceleration, $$a$$, would then depend on of this velocity since we have by linear acceleration,

$$1400*2{\pi} = \frac{1}{2}a 40^{2} + v*40$$

Obviously we see that the acceleration is dependent on the choice of the initial velocity.

where did the 1400*2pie come from??

 

[Werg22]

Ok, hage, I have to apologize sincerely. Your solution was right, and I was completely wrong. I had disregarded that the fact that it comes to a stop dictates the solution. Sorry buddy, my bad. xXmarkXx, just set two equations.

$$0 = v_{i} + 40a$$

and

$$1200*2{\pi} = \frac{1}{2}a40^{2} + 40v_{i}$$

You have two equations and two variables, and you solve. The 1200*2pi comes from the fact that at each revolution, the disk covers 2pi radians. So for 1200 revolutions, it will cover 1200*2pi radians. Once again, sorry hage, I got cocky and stupid.

 

[xXmarkXx]

right, i end up with the angular acceleration to be just pie.. v_i=-40a, then i plugged that into the other equation and solved for a. which is pie. Is that what you guys got??

 

[hage567]

No hard feelings, Werg22.

 

[xXmarkXx]

and now i used pie=a and solved for the angular velocity, which i had gotten before to be equal to 40*pie...agreed??

 

[xXmarkXx]

my fault, i checked it over again and now i got -3pie = angular acceleration. Would you guys at least tell me if i am on the right track or not?

 

[hage567]

my fault, i checked it over again and now i got -3pie = angular acceleration. Would you guys at least tell me if i am on the right track or not?

Yes, you appear to be on the right track.

 

[xXmarkXx]

and then i got 120pie = w₀
how bout that? correct o mundo? Thanks hage and werg for the help and being patient with me. I'm a slow learner.

 

[hage567]

Looks good to me.

You're welcome.