Angular Velocity Differential Equation

[ThomasMagnus]

Problem:

Newton's 2nd Law for rotational motion states that the product of the moment of Inertia, I, and the angular acceleration alpha(t) is equal to the net Torque acting on a rotating body. Consider a wheel that is being turned by a motor that exerts a constant torque T. Friction provides an opposing torque proportional to the square root of the angular velocity w(t). The moment of inertia of the wheel is I.

A) Write out a differential equation that represents Newton's 2nd law for rotational motion in this case. Note that alpha(t)=w'(t)

B) If the initial angular velocity is w_o, determine the angular velocity as a function of time, w(t). Can you find an expression for w(t)

Attempt at Solution:

I was able to find the answer for A. I found that it was: I(dw/dt)=T-k(sqrt(w))

As for part B, I am really confused. I know I have to separate the equation and integrate it. I know that dw=w'(t)dt.Other than that I'm lost. First of all, should I be plugging in Wo now or at the end? I'm trying to isolate the w's but I keep getting:
I *w'(t)+k(sqrt(w))=T

I think I need to find a way to get the w's as a product or quotient of each other, so I can use dw=w'(t)dt.

 

[Dick]

Problem:

Newton's 2nd Law for rotational motion states that the product of the moment of Inertia, I, and the angular acceleration alpha(t) is equal to the net Torque acting on a rotating body. Consider a wheel that is being turned by a motor that exerts a constant torque T. Friction provides an opposing torque proportional to the square root of the angular velocity w(t). The moment of inertia of the wheel is I.

A) Write out a differential equation that represents Newton's 2nd law for rotational motion in this case. Note that alpha(t)=w'(t)

B) If the initial angular velocity is w_o, determine the angular velocity as a function of time, w(t). Can you find an expression for w(t)

Attempt at Solution:

I was able to find the answer for A. I found that it was: I(dw/dt)=T-k(sqrt(w))

As for part B, I am really confused. I know I have to separate the equation and integrate it. I know that dw=w'(t)dt.Other than that I'm lost. First of all, should I be plugging in Wo now or at the end? I'm trying to isolate the w's but I keep getting:
I *w'(t)+k(sqrt(w))=T

I think I need to find a way to get the w's as a product or quotient of each other, so I can use dw=w'(t)dt.

It's a separable differential equation. I(dw/dt)=T-k(sqrt(w)) so dw/(T-k(sqrt(w))=dt/I. Integrate both sides. It's a differential equations thing. The w0 part is after you've got the general solution.

 

[ThomasMagnus]

Ok, now I see it! However, I get: int (dw/T-Ksqrt(w)) on one side. I tried subbing for the whole denominator, but it doesn't seem to work..

 

[ThomasMagnus]

I \frac{dω}{dt} = τ-k\sqrt{w}
dt(τ-k\sqrt{w})=(I)dω
\frac{dt}{I} = \frac{dω}{τ-\sqrt{w}}
\int \frac{dt}{I}= \int \frac{dω}{τ-k\sqrt{w}}
\frac{1}{I}\int dt= \int \frac{dω}{τ-k\sqrt{w}} (I is constant so you can do this I believe?)
\frac{t}{I}=\int \frac{dω}{τ-k\sqrt{w}}
\int \frac{dω}{τ-k\sqrt{w}} u=τ-k\sqrt{ω}, du=-\frac{k}{2\sqrt{ω}} dω
\frac{1}{k}2\sqrt{w}du=dω, (\frac{1}{k})(-u+τ)=\sqrt{ω} so dω=(\frac{1}{k})2(-u+τ)du=\frac{-2u+2τ}{k}du
\int \frac{dω}{τ-\sqrt{w}} = \int \frac{-2u+2τ}{ku} du
= \frac{1}{k}\int\frac{-2u}{u}du + \frac{1}{k}\int \frac{2τ}{u}du
= \frac{-2}{k}\int du + \frac{2τ}{k}\int \frac{1}{u} = -2\frac{u}{k}+\frac{2τ}{k}*ln|u|+c
= -2(τ-k\sqrt{w})\frac{1}{k} + 2\frac{τ}{k} *ln|τ-k\sqrt{ω|}+c
\frac{-2}{k} (τ-k\sqrt{ω}+τln|τ-k\sqrt{ω}|)+c = \frac{t}{I}

I'm lost at this point. I think everything is OK up to here? How would I plug ω_o in? Would I plug it in so that \sqrt{ω} becomes \sqrt{w-wo} which is like \sqrt{\Deltaω} ? or ω(0)=ω_o ?
My solutions page says it should be k(√ω-√ω_o)+τ ln \frac{τ-k\sqrt{w}}{τ-k\sqrt{ωo}}= -k²t/2I

 

[Dick]

I \frac{dω}{dt} = τ-k\sqrt{w}
dt(τ-k\sqrt{w})=(I)dω
\frac{dt}{I} = \frac{dω}{τ-\sqrt{w}}
\int \frac{dt}{I}= \int \frac{dω}{τ-k\sqrt{w}}
\frac{1}{I}\int dt= \int \frac{dω}{τ-k\sqrt{w}} (I is constant so you can do this I believe?)
\frac{t}{I}=\int \frac{dω}{τ-k\sqrt{w}}
\int \frac{dω}{τ-k\sqrt{w}} u=τ-k\sqrt{ω}, du=-\frac{k}{2\sqrt{ω}} dω
\frac{1}{k}2\sqrt{w}du=dω, (\frac{1}{k})(-u+τ)=\sqrt{ω} so dω=(\frac{1}{k})2(-u+τ)du=\frac{-2u+2τ}{k}du
\int \frac{dω}{τ-\sqrt{w}} = \int \frac{-2u+2τ}{ku} du
= \frac{1}{k}\int\frac{-2u}{u}du + \frac{1}{k}\int \frac{2τ}{u}du
= \frac{-2}{k}\int du + \frac{2τ}{k}\int \frac{1}{u} = -2\frac{u}{k}+\frac{2τ}{k}*ln|u|+c
= -2(τ-k\sqrt{w})\frac{1}{k} + 2\frac{τ}{k} *ln|τ-k\sqrt{ω|}+c
\frac{-2}{k} (τ-k\sqrt{ω}+τln|τ-k\sqrt{ω}|)+c = \frac{t}{I}

I'm lost at this point. I think everything is OK up to here? How would I plug ω_o in? Would I plug it in so that \sqrt{ω} becomes \sqrt{w-wo} which is like \sqrt{\Deltaω} ? or ω(0)=ω_o ?
My solutions page says it should be k(√ω-√ω_o)+τ ln \frac{τ-k\sqrt{w}}{τ-k\sqrt{ωo}}= -k²t/2I

You are almost right. I think you dropped a factor of k and a sign. Check it. To get the constant c put t=0, and $\omega(0)=\omega_0$. Once you've solve for c put it back into the solution and simplify.