Doubt concerning centripetal acceleration Question 1 :
What is the centripetal acceleration and angular velocity of a child located $8.2 \ \mathrm{m}$
the center of a carousel? The speed (size of the tangential velocity) of the child is
$2.1 \ \mathrm{m / s}$.
Question 2 :
A train moves in a straight path north until it turns to west.
If the road segment used to change direction is shaped like a quarter circle of radius $30 \ \mathrm{m}$ and the train takes $30 \ \mathrm{s}$ to traverse that part of the road,
What is it the speed (size of the velocity vector) and the centripetal acceleration acts on the train as it traverses the curve?
I am reviewing some concepts like centripetal  force, $a_r = ( v^2 ) / r$
also this :

The direction of the centripital
acceleration is always inwards along the radius vector of the circular motion. The magnitude of the centripetal acceleration is related to the tangential speed and angular velocity as follows :
$$a_c =\frac {v_t ^2}{r} = \omega ^2 r$$

Can you please guide me to solve the 2 problems above?
For the fisrt one is it only  $(2.1 \  \mathrm{m / s)^2 / \  8.2 \ m}$  ?
 A: cMinor.  You are good to go on the first problem.  Just crank through the math.  You should get an answer just over 0.5 (m/(s^2)).
On the train problem, find the path length of the quarter circle, then divide it by 30 seconds to get train speed.  You will then have all you need to do the (v^2)/r thing for that problem.
Angular velocity does not need to be found to solve either of these problems.
A: On this class of problems, there's a secret that allows you to get the answer without having to memorize any formulas. (By the way, I got a 990 on the physics GRE so you can trust that I use this sort of thing.)
You have three things that contribute to the problem: $m$, $r$ and $v$. These have three different units, $[kg], [m], [m/s]$. Therefore, anything you want can be obtained from these in only one way.
Centripetal acceleration is in units of $[m/s^2]$, correct? from this we see that mass does not enter into it. We have velocity $[m/s]$ and radius $[m]$ and the only way we can get $[m/s^2]$ from these is by $v^2/r$ which is what you wrote.
The one caveat on using this trick is that you want to avoid things associated with the period because you'll need a factor of $2\pi$. For example, instead of computing the frequency in Hz or $[1/s]$, you will get the angular frequency (i.e. the rate at which the object eats up angles). To get this sort of thing, just recall that the distance around the circle is $2\pi r$.
