A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 4 sec .

[manal950]

A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 4 sec .
The shaft continues to accelerate at this rate for the next 8 sec . Thereafter the shaft rotates with a uniform angular speed . Find the total time to complete 400 rev .

my answer is ...

10 rev/s----------------18rev/s-------------------
4s
W0 = 62.83 rad/s
Wf = 113.0
total time at 400 rev ========400/2Pi
so Q = 2513 rad
w = w0 + at
133.0 = 62.83 + at
a = 12.4 so this is niforml accelerated
now I will take the velocity at 12 sec (4+8 = 12 )
so
w = w0 + w
w = 113.0 + 12.4 X (12)
=261.8

------------ now I will take time after 12 s which the a = 0 because of angular speed ----

Q = w0 +1/2 at^2
2513=261.8t+1/2(0)^2 t
t= 9.5 s

so Total time = 12 + 9.5 = 21.5 s



t = 2513/261.8 =

 

[collinsmark]

A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 4 sec .
The shaft continues to accelerate at this rate for the next 8 sec . Thereafter the shaft rotates with a uniform angular speed . Find the total time to complete 400 rev .

my answer is ...

10 rev/s----------------18rev/s-------------------
4s
W0 = 62.83 rad/s
Wf = 113.0
total time at 400 rev ========400/2Pi
so Q = 2513 rad
w = w0 + at
133.0 = 62.83 + at

I think you mean "113.0" Not "133.0." I'm guessing that's just a typo.

a = 12.4 so this is niforml accelerated

Even with using "113.0" your calculation is off by a significant figure. See my hint below on how to be more exact on some of these calculations.

now I will take the velocity at 12 sec (4+8 = 12 )
so
w = w0 + w
w = 113.0 + 12.4 X (12)

If you are looking for the final velocity after t = 12 seconds using that formula, you need to start with initial velocity at 12 seconds previous (t = 0). You used the initial velocity figure at 8 seconds previous (i.e. t = 4 s).

Here's a hint that might help avoid rounding errors as you solve this problem. Keep things in terms of π.

For example, when you convert the initial angular velocity from 10 rev/s to units of radians per second, leave it as ω_i = 20π rad/s. Similarly, when solving for the acceleration, leave things in terms of π.

(36π [rad/s] - 20π [rad/s])/(4 ) = 4π [rad/s²]​

You can convert the π to its numerical value at the very end, if it doesn't happen to completely cancel out before then. In the mean time, you'll have to carry πs around, but you won't have to worry about any rounding errors, and you might not even need to use a calculator at all!

[Edit: another potential option is to stick with revolutions instead of radians all along. As long as you never need to convert to linear distance/velocity/acceleration, it will still work.]

 

[manal950]

thanks sir ,, but know you don't show to me where I do the mistake and

 

[collinsmark]

thanks sir ,, but know you don't show to me where I do the mistake and

I'll try again, and highlight in red.

W0 = 62.83 rad/s
Wf = 113.0
total time at 400 rev ========400/2Pi
so Q = 2513 rad
w = w0 + at
133.0 = 62.83 + at

I think you meant 113.0 [rad/s], not 133.0. I'm guessing that's just a typo though.

a = 12.4 so this is niforml accelerated

There's a rounding error involved with that. The exact angular acceleration is 4π [rad/s²] ≈ 12.56637 [rad/s²]

now I will take the velocity at 12 sec (4+8 = 12 )
so
w = w0 + w
w = 113.0 + 12.4 X (12)

The 113.0 [rad/s] value was calculated at t = 4.0 . Yet your equation assumes that the acceleration continues for 12 after that. Either you need to use the ω₀ = 62.83 [rad/s] *OR* change Δt to 8 . (Either one will work -- just do one or the other, not both.)

As far as my previous hint goes, if you keep 'π' as a symbol, rather than a number, it can help avoid rounding errors, if you wish to work everything in units of rad, rad/s, and rad/s².

And there's nothing wrong with converting everything to rad, rad/s, and rad/s² if you are more comfortable working that way. But for this particular problem, it's actually not necessary. You can work directly with θ in units of rev, ω in units of rev/s, and α in units of rev/s^s, if you want to. The choice is yours, but just be sure to keep track of your units and stay consistent, whichever way you choose.

 

[manal950]

Ok I will do next time the anwer with pi ... but let me complete in this Q ^_^

W0 = 62.83 rad/s
Wf = 113.0
total time at 400 rev ========400/2Pi
so Q = 2513 rad
w = w0 + at
113.0= 62.83 + a(4)
a = 12.5 rad/s^2

now I will take the velocity at 12 sec (4+8 = 12 )
so
w = w0 + w
w = 62.8+ 12.4 X (12) (12)
w = 213.13 rad/s

Q = w0t +1/2 at^2
2513=12.5t+1/2(0)^2 t
t= 11 s

so Total time = 12 + 11 =
so total time now I got 23.7 s

please check my answer now ...

 

[collinsmark]

Ok I will do next time the anwer with pi ... but let me complete in this Q ^_^

That's fine.

W0 = 62.83 rad/s
Wf = 113.0
total time at 400 rev ========400/2Pi
so Q = 2513 rad
w = w0 + at
113.0= 62.83 + a(4)
a = 12.5 rad/s^2

'Looks good!

now I will take the velocity at 12 sec (4+8 = 12 )
so
w = w0 + w
w = 62.8+ 12.4 X (12) (12)
w = 213.13 rad/s

That's roughly what I got (to the first two or three significant figures anyway).

Q = w0t +1/2 at^2
2513=12.5t+1/2(0)^2 t
t= 11 s

That's not quite right. I see where the 2513 comes from. That's

3513 [rad] ≈ (2π [rad/rev])(400 [rev])​

so that's the total number of radians (2π revolutions) that the problem statement is asking you to find.

But you've calculated the time it takes to make 3513 [rad] at a constant angular velocity of 12.5 [rad/s].

A couple of things:

1. The angular velocity is never at a constant 12.5 [rad/s]. You seem to be confusing angular velocity with the 12.5 [rad/s²] angular acceleration.
2. Not all of the total 400 rev (~3513 [rad]) happen at a constant angular velocity. A large part of them happen when the shaft is accelerating, and only the rest happen at a constant angular velocity.

Here's another hint: What's the angular displacement (i.e "how many radians are turned") after the first 12 seconds, when the shaft is accelerating? (i.e. what is θ at t = 12 [sec], knowing that the shaft accelerates from time 0 < t < 12 [sec])

 

[manal950]

sorry little i have confusing anyway

Q = w0 + 1/2at^2
Q = 213.13 + 1/2(12.5)X(12)^2
Q = 1113.3 rad

now what ... ?

 

[collinsmark]

sorry little i have confusing anyway

Q = w0 + 1/2at^2
Q = 213.13 + 1/2(12.5)X(12)^2
Q = 1113.3 rad

now what ... ?

You forgot one of the 't's in the formula. See in red below.

θ = θ₀ + ω₀t + ½αt²

Also, 213.13 [rad/sec] is the final angular velocity that you calculated earlier, corresponding to the end of acceleration (at t = 12 [sec]). The formula above calls for the initial angular velocity at the moment the acceleration began (at t = 0).

 

[manal950]

Q = 3457.56

Now what ?

 

[collinsmark]

Q = 3457.56

Now what ?

Try that one again.

Q = w0 + 1/2at^2
Q = 213.13 + 1/2(12.5)X(12)^2
Q = 1113.3 rad

ω₀ ≠ 213.13 [rad/s]

213.13 [rad/s] is the final angular velocity, not the initial.

 

[manal950]

sorry ...

Q = 62.83 + 1/2(12.5)X(12)^2
= 962.83 rad

Now wha

 

[collinsmark]

sorry ...

Q = 62.83 + 1/2(12.5)X(12)^2
= 962.83 rad

'Forgot the t again. (θ = θ₀ + ω₀t + ½αt²)

 

[manal950]

Q = 1653.96 rad
we want get Q at 12 s what you mean by Q = Q0 + w0t+1/2t^2
what is Q0 ??

 

[collinsmark]

Q = 1653.96 rad

Allow me to start with a summary of what we've done so far.

You've determined the acceleration, using the interval from 0 < t < 4 .

ω_f = ω_i + αt

18 [rev/s] = 10 [rev/s] + α(4 )

8 [rev/s] = α(4 )

(8 [rev/s])/(4 ) = α

α = 2 [rev/s²]

(Although it's not necessary, you could convert that to radians per second squared by multiplying by 2π [rad/rev], such that (2π [rad/rev])(2 [rev/s²]) ≈ 12.566 [rad/s²)

You've calculated the final angular velocity, after accelerating for 12 .

ω_12s = ω_i + αt

= 10 [rev/s] + (2 [rev/s²])(12 )

= 10 [rev/s] + 24 [rev/s]

ω_12s = 34 [rev/s]

(Although it's not necessary, you could convert that to radians per second by multiplying by 2π [rad/rev], such that (2π [rad/rev])(34 [rev/s]) ≈ 213.628 [rad/s])

And you've calculated the angular displacement at t = 12 .

θ_12s = θ₀ + ω_it + ½αt²

= 0 + (10 [rev/s])(12 ) + ½(2 [rev/s²])(12 )²

= 120 [rev] + 144 [rev]

θ_12s = 264 [rev].

(And again, although not necessary, that's approximately 1658.76 [rad/sec])

Now wha

The first 12 seconds used up 264 revolutions. The problem statement asks for 400 revolutions, so how many revolutions are left to go?

You know that the remaining revolutions take place at the final angular velocity of ω_12s. So how long does it take to rotate through the remaining revolutions at a constant angular velocity of ω_12s?

 

[manal950]

thaaaaaaaaaaaaaaaaaaaaaaanks good summary that useful for me

Q = 400 - 260 = 140 revolutions

 

[collinsmark]

thaaaaaaaaaaaaaaaaaaaaaaanks good summary that useful for me

Q = 400 - 260 = 140 revolutions

Hold on, θ_12s = 264 [rev], not 260.

After making that correction, you have one (or maybe two) more calculation(s) to go and you're finished!

 

[azizlwl]

Draw a revolutions/sec- time diagram.
You have a line with a point (a sec, 10rev/sec) and another point at (a+4sec, 18rev/sec)
Since this is a constant acceleration, the gradient of the line is the acceleration.

Now continue the line until time=(a+4+8)sec. This point is ((a+4+8)sec, X rev/sec)
You should be able to find the value of X

As you should know the area under the line and time axis is the revolutions covered within that period of time

If less than 400rev add then another area(rev) with constant rev/sec.
The value of this constant velocity is equal to X as in para. 2.

 

[manal950]

In fact I don't know which correction you mean now from 0 to 12 we have 264 revolutions so this the displacement so remaining revolutions Q = ( 400 - 264 ) = 136 rev

 

[collinsmark]

In fact I don't know which correction you mean now from 0 to 12 we have 264 revolutions so this the displacement so remaining revolutions Q = ( 400 - 264 ) = 136 rev

Yes, that's right!

Almost there.

How long does it take a shaft to rotate 136 revolutions, if it is rotating at the constant velocity ω_12s? (You've already calculated what ω_12s is, btw.)

 

[manal950]

that why I face problem in like this Q because I don't know how to draw
just I do the period as a line like this
4s--------------12s-----------------ts(400rev)
a = 12.5-----w =213.623
w0 =13-------Q = 1658.76

---------anyway ... I complete my answer ---------

How long does it take a shaft to rotate 136 revolutions
Q = 854.8 rad
w = 213.623
t = 12
a = 0

so after that I use this two equation ..
Q = W0 + 2at^2
136 = W0(12) + 2 X (0) ( 12^2)
w0 = 11.3 rad/s

from this Q we get time at 136 rev ...

w = w0 +at
t = 213.62 - 11.3/0 OMG ... sure is wrong !>>

 

[azizlwl]

This a sample of y-x diagram.
In your case replace y with ω or rec/sec and x with second.
Then mark the points as my previous post.
Make a straight line adjoining point #1 and point #2.
Make a vertical line at point example 1 second.
Second vertical line at 1+4 sec
Third line at 1+4+8 sec.
You can calculate the area below the straight line which equal to rev covered.
Hope by the drawing you can visualize what's happening to the object.

As you can see, you can derive all the equations you have mentioned from the diagram.

 

[manal950]

http://store2.up-00.com/May12/hKh28157.jpg

 

[azizlwl]

Good you try.
Only the data should be of the given ones.

A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 4 sec .
The shaft continues to accelerate at this rate for the next 8 sec . Thereafter the shaft rotates with a uniform angular speed . Find the total time to complete 400 rev

ω0 must be 10 rev/sec
ω1 = 18 rev/sec
ω2=? (just continue straight line from ω0,ω1 to intersection of time t=12sec)
Then uniform angular speed ω2(horizontal line from (ω2,t2) to the right.

t0=0 sec
t1=4 sec
t2=12 sec
t3=?

 

[manal950]

Now is ok ...

http://store2.up-00.com/May12/ljA30374.jpg

 

[azizlwl]

No

The time is correct with value.
You do the same with vertical axis.
Mark where 10 rev/sec and 18rev/sec. Not ω0 and ω1

 

[manal950]

can please explain to me more ?

 

[azizlwl]

Instead of w2 replace it with 10 rev/sec
Instead of w1 replace it with 18 rev/sec
Lastly replace w0 with w2.
This is because you know the values of w0 and w1
The only unknown is w2.

The order should be the lowest w0 follow by w1 and w2

 

[collinsmark]

How long does it take a shaft to rotate 136 revolutions
Q = 854.8 rad
w = 213.623
t = 12
a = 0

so after that I use this two equation ..
Q = W0 + 2at^2
136 = W0(12) + 2 X (0) ( 12^2)
w0 = 11.3 rad/s

You're supposed to be solving for the additional time t here, not angular velocity. You already know what ω₀ is: 213.626 [rad/s] (same thing as 34 [rev/s]).

Also, the '136' figure is in units of revolutions. If you want to work in radians, that's fine, but you'll have to convert that value to radians.

But your

θ = ω₀t + ½αt²

with α = 0, is good. Solve for t.

But make sure you are consistent with your units. You'll notice that in all my posts to this thread I've put in the efforts to carry all the units though. I did that for a reason. Carrying the units though helps avoid mistakes in inconsistency.

 

[azizlwl]

http://img607.imageshack.us/img607/1910/p1010014xk.jpg