Here's the problem with my attempts (some of the answers seem really fishy):
"A centrifuge starts from rest on a curve with a radius of 130 m and has a constant tangential acceleration of $1.4 m/s^2$."
1) Find the time it takes for the car to increase its total acceleration to $2.9 m/s^2$.
We know the square root of the sum of $a_t$ and $a_r$ (for tangential and radial acceleration, respectively) equals total acceleration. w is angular velocity and $\alpha$ is angular acceleration.
$ (1.4^2+a_r^2)^{0.5}=2.9\\ a_r=2.54 m/s^2 \\ a_r= w^2/r\\ w^2=2.54*130 w=18.17\ rad/s\\ -->w = \alpha*t_f\\ 18.17=0.011*t_f\\ t_f=1651.8s. $
Is this correct? I can't imagine it would take this long...tbh
2) What is the centrifuge's tangential velocity after this time interval?
$v_t=w*r 18.17*130=2352.1 m/s$
3) What angle does the centrifuge turn during this time interval
$\theta =0.5 \alpha t_f^{2}$ 0.5(0.011)(1651.8)=9.085 radians$
Can some of you please comment on whether I'm using the formulas correctly and troubleshoot my understanding here? I would really appreciate the help.
$a_t=1.4\,\mathrm{m s}^{-2}$, $a=2.9\,\mathrm{m s}^{-2}$, $r=130\mathrm{m}$
$$\begin{align} \sqrt{a^2-a_t^2}&=a_r=2.54\,\mathrm{m s}^{-2}\\ a_r&=\frac{v^2}{r}\quad \Longrightarrow \quad v=\sqrt{a_r r}=18.17\,\mathrm{m s}^{-1}\\ v&=a_t\,t\quad \Longrightarrow \quad t=\frac{v}{a_t}=12.98 \,\mathrm{s}\\ \theta&=\frac{1}{2}\alpha t^2=\frac{1}{2}\frac{a_t}{r}t^2=0.91\,\mathrm{rad} \end{align} $$