Angular velocity by time variable force

[Totally]

Homework Statement

Force represented by F(t)=20+t+5t^2 [N] acts upon the rim of a disk. How much time has to pass before the disk has angular velocity of 200 revs\sec? R and I are known.

Homework Equations

\begin{equation*}
\tau=I\alpha
\end{equation*}
\begin{equation*}
\tau=R \times F
\end{equation*}
\begin{equation*}
I=\frac{1}{2}mR^2
\end{equation*}

The Attempt at a Solution

\begin{equation*}
\alpha=\frac{R}{I}F
\end{equation*}
\begin{equation*}
\omega=\frac{R}{I}\int_{0}^{t} F(t)dt
\end{equation*}
I have same variable in my upper integration limit as in the function. So I'm thinking of integrating F(t) twice to get the distance and instead of writing it as θ, write it as ωt so it looks like the following
\begin{equation*}
\omega t=\frac{R}{I}∫∫F(t)dt
\end{equation*}
Then divide by t and solve the cubic. Does that seem like a legitimate move?

 

[Chestermiller]

It seems legitimate for getting ω, but expressing θ as ωt is not correct. However, if θ were put in place of ωt on the left side of your final equation, the equation would be correct. However, the problem is not asking for θ. It is just asking for the time t at which ω reaches a certain value. This can be determined exclusively with your first equation for ω.

Chet

 

[Totally]

Oh wow, I can just integrate F(t) as indefinite integral, can't I? I don't need the 0 and t limits because the integration constant I'll be getting will mean initial velocity, which is zero in this case.

Also, why is it wrong mathematically to sub θ for ωt? Is it because I'm supposed to be integrating t on the right side?

 

[Chestermiller]

Oh wow, I can just integrate F(t) as indefinite integral, can't I? I don't need the 0 and t limits because the integration constant I'll be getting will mean initial velocity, which is zero in this case.

I would write the equation as
$$\omega=\frac{R}{I}\int_{0}^{t} F(t')dt'$$
where t' is a dummy variable of integration. This is actually a definite integral that gives you the value of ω at t'=t.

Also, why is it wrong mathematically to sub θ for ωt? Is it because I'm supposed to be integrating t on the right side?

θ is not equal to ωt unless ω is constant. Otherwise, $θ=\int_0^t{ω(t')dt'}$, assuming that θ=0 at t = 0.

Chet