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Angular displacement is represented by the symbol $\theta.$ Angular velocity is the rate of change of angle and is denoted by $\omega$ or $\overset{.}{\theta}. $ Angular acceleration, which is the rate of change of angular velocity, is denoted by $\overset{..}{\theta},\ $ or $\ \alpha,$ as we now consider angular acceleration to be constant.

My book then derives/obtains the following equation: $\omega_1 = \omega_0 + \alpha t,\ $ and I am vaguely OK with the derivation, although I personally find it non-rigorous and handwavy.

Then it asks to:

Show by integration, that:

$$ \theta = \omega_0 t + \frac{1}{2} \alpha t^2.$$

My attempt:

$$\int_{0}^{t}\omega_1\ dt = \int_{0}^{t} \omega_0 + \alpha t\ dt \implies \omega_1 t= \omega_0 t + \frac{1}{2} \alpha t^2. $$

This suggests that $\omega_1 t = \theta,\ $ which doesn't make sense to me. What is going on? I think there is some speed = distance / time thing but I'm confused as to how it is being applied here.

Adam Rubinson
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    $w_0$ is a constant but $w_1$ depends on $t$ so I do not think you can evaluate the first integral as $w_1t$. – Vasili Jul 28 '22 at 13:19

1 Answers1

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I will address why you have to integrate $\omega_1$ with respect to $t$:

$\omega_1$ is a function of $t$; and $$\omega_1(t)=\dot\theta=\frac{d\theta}{dt}$$ If $\omega_1$ were constant, then your reasoning of the integral $\displaystyle\int_0^t \omega_1(t)dt$ being equal to $\omega_1t$ would be correct. However, that is not the case, as $\omega_1$ clearly varies with $t$.

Thus we separate the differential equation and write : $$\int\omega_1(t)dt=\int d\theta$$ so that we get $$\theta=\int( \omega_0+\alpha t )dt$$$$=\omega_0 t+\frac12 \alpha t^2+C.$$ Now we have to find $C$ so we use the initial values. In the problem it should be given that $\theta$ at time $t=0$ is equal to $0$ (which you seem to have assumed). Otherwise, for a more general case, if we take the initial angular displacement at time $t=0$ to be $\theta_0$, we get $$\theta= \theta_0+\omega_0 t+\frac12 \alpha t^2.$$

POSTSCRIPT: If you want to write the integral $\displaystyle\int_0^t\omega_1 t dt$ as $c\cdot t$, where $c$ is some constant, then $c$ must be the time average of $\boldsymbol\omega_1$ during the time interval $0-t$, i.e, $\displaystyle\int_0^t \omega_1(t)dt=\langle\omega_1(t)\rangle t$, where $\langle\omega_1(t)\rangle$ is the time average.

insipidintegrator
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  • So the point is that: $\theta(t=0) = 0.\quad \theta$ ( or $\theta(t=t)$ ) is the final angular displacement, and $\omega_1$ is the final angular velocity. Therefore $\omega_1 = \frac{d\theta}{dt}.$ – Adam Rubinson Jul 28 '22 at 13:56
  • $\theta$ IS the final angular displacement. 2. $\omega_1$ is NOT the final angular velocity. It is a variable; it is the angular velocity for ANY time t, given by $\omega_1=\omega_0+\alpha t$. 3. $\theta(0)=0$ must be assumed for the formula in the problem to hold valid.
    • – insipidintegrator Jul 28 '22 at 14:20
  • If $\omega_1$ is the angular velocity at any time $t$, then what is $\omega?$ Are $\omega_1$ and $\omega$ the same thing? – Adam Rubinson Jul 28 '22 at 16:01
  • Oh, sorry I’ll change the variable to avoid confusion. – insipidintegrator Jul 28 '22 at 16:02
  • The book says that $\omega_1$ is the final (angular) velocity. So I think the book is viewing the final (angular) velocity as a variable. That would make some sense to me... or is this incoherent? – Adam Rubinson Jul 28 '22 at 16:16
  • I mean, you could say that, if initial point is t=0 and final point is t=$t$, then $\omega_1$ could be viewed as the final angular velocity. – insipidintegrator Jul 28 '22 at 16:20
  • Yes I think that is the way the book is viewing it and is also the way that makes sense to me. Thanks for your answer though. It did help – Adam Rubinson Jul 28 '22 at 16:24