My Math Forum A shaft is uniformly accelerated from 10rev/s to 18 rev/s in

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 May 24th, 2012, 09:58 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 A shaft is uniformly accelerated from 10rev/s to 18 rev/s in 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 . [color=#FF40BF]my answer is ...[/color] 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 =
 May 24th, 2012, 10:45 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: A shaft is uniformly accelerated from 10rev/s to 18 rev/ We are given: $\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=4\pi\tex t{ \frac{rad}{s^2}}$ where $\omega_0=20\pi\text{ \frac{rad}{s}},\,\theta_0=0\text{ rad}$ for $0\le t\le12$ Hence: $\theta(t)=2\pi t^2+20\pi t=2\pi$$t^2+10t$$$ Thus, the number of revolutions in the first 12 second is: $\frac{\theta(12)}{2\pi}=264$ The angular velocity at $t=12$ is: $\omega(t)=at+\omega_0$ hence: $\omega(12)=4\pi\cdot12+20\pi=68\pi$ Now, for $12 we have: $\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=0\text{ \frac{rad}{s^2}}$ where $\omega_{12}=68\pi\text{ \frac{rad}{s}},\,\theta_{12}=528\pi\text{ rad}$ Hence: $\theta(t)=68\pi t+528\pi$ Setting $\theta(t)=800\pi$ we find: $800\pi=68\pi t+528\pi$ $800=68t+528$ $t=4$ Thus, the total time in seconds is: $t=12+4=16$
 May 24th, 2012, 02:38 PM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A shaft is uniformly accelerated from 10rev/s to 18 rev/ Markfl sorry ..I have confusing ,, why solve the Q in this way why you don't use this formula for rotating motion w = w0 + at w^2 = wo^2 + 2 a Q Q = ut + 1/2Q t^2 which I did ?
 May 25th, 2012, 09:29 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: A shaft is uniformly accelerated from 10rev/s to 18 rev/ If you'll notice, I used/derived the same formulas, I just used the variables I was taught to use: ? for the angle ? for the angular velocity So, just substitute for these the variables you like.
 May 26th, 2012, 01:14 PM #5 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A shaft is uniformly accelerated from 10rev/s to 18 rev/ Hi the teacher said here t = 16 s
 May 26th, 2012, 03:32 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: A shaft is uniformly accelerated from 10rev/s to 18 rev/ What is your teacher's reasoning? edit: I found my error and corrected my post above.
 May 26th, 2012, 06:02 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: A shaft is uniformly accelerated from 10rev/s to 18 rev/ We need not convert from revolutions to radians. We are given: $\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=\alpha=\ frac{18-10}{4}\:\text{\frac{rev}{s^2}}=2\text{ \frac{rev}{s^2}}$ where $\omega_0=10\text{ \frac{rev}{s}},\,\theta_0=0\text{ rev}$ for $0\le t\le12$ Hence: $\theta(t)=$$t^2+10t$$\text{ rev}$ Thus, the number of revolutions in the first 12 seconds is: $\theta(12)=264\text{ rev}$ The angular velocity at $t=12$ is: $\omega(t)=\alpha t+\omega_0$ hence: $\omega(12)=2\cdot12+10=34\text{ \frac{rev}{s}}$ Now, for $12 we have: $\frac{d^2\theta}{dt^2}=\frac{d\omega}{dt}=0\text{ \frac{rev}{s^2}}$ where $\omega_{12}=34\text{ \frac{rev}{s}},\,\theta_{12}=264\text{ rev}$ Hence: $\theta(t)=34t+264$ Setting $\theta(t)=400\text{ rev}$ we find: $400=34t+264$ $t=4$ Thus, the total time in seconds is: $t=12+4=16$

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