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.