Calculate the constant angular acceleration if 3600 revolutions are made in two minutes starting from rest.

Question

I am currently working through Morris Klines "Calculus: An intuitive approach" and I am struggling with a particular question: Calculate the constant angular acceleration if 3600 revolutions are made in two minutes starting from rest.

I understand how to calculate derivatives by iteration, I don't want a simple answer but an explanation that may help for me to understand how to get the answer.

I know that distance along a rotation is =$R\theta$, and angluar velocity is =$R\theta'$ and angular rotation is $R\theta''$.

I've calculated the rotation in $rads/sec$ as 1.5 pi.

What do I need to consider to derive the correct answer (which is pi $rads/sec^2$)

Many thanks.

Answer 1

$3600$ revolutions in $2$ minutes is an average speed of $30$ revolutions per second, which is $60 \pi$ radians per second. Just as with constant linear acceleration, the final speed starting from rest will be twice the average speed - because

$\displaystyle \frac s t = \frac {\int v dt}{t} = \frac {\int at dt}{t} = \frac 1 2 at$

So the final speed is $120 \pi$ radians per second. If you divide this by the time ($120$ seconds) then you get the acceleration.

Answer 2

$3600$ revs are made in $120$ seconds starting from rest. Let the angular acceleration be $\alpha$ revs/second$^2$. In $t$ seconds the angular speed gained would be $= \alpha t$ revs/second. Also the angular distance covered during any interval $dt$ would be angular speed at that time multiplied with $dt$. Hence integrating from $0$ to $120$ we get Angular distance = $\int_0^{120} \alpha t dt = 3600$ revs.

$\implies \alpha = 0.5$ revs/ second$^2$