[![Figure][1]][1]

> The crank $OA$ is rotating around $O$ in plane of the figure with

> constant angular velocity. $OA$ is making connecting rod $AB$ move the

> circle along the arc with no slip. Given radius of the arc $R = 1$m,

> radius of the circle $r = 0.5$m and angular velocity $ω = 2$

> rad/sec = const, and $OA = AB = R$. Find velocity (tangential) and acceleration of

> point A and point B at the presented moment and the location of instant centre of acceleration.

I'm stuck finding the acceleration of B.

I found the velocity of points using $v*{B} = v*{A} = ω*r = ω*OA = 2$ m/s. I also found out that K is the instant centre of velocity of the circle.

Since $ω = const$ then $α = 0$ and the acceleration of A equals $ω^2*OA = 4 m/c^2$. Because the velocities of A and B are equally directed then $ω_{AB} = 0$.

$a*{B} = a*{B}^t + a*{B}^n$ (tangential and normal acc.) as B is rotating around Q. B is also the part of AB, then $a*{B} = a*{A} + a*{AB}^t + a_{AB}^n$.

[![enter image description here][2]][2]

[1]: https://i.stack.imgur.com/oCvLm.jpg

## [2]: https://i.stack.imgur.com/kOan5.jpg

Now lets equate and project theese accelerations on x and y

x: $-a*{B}^t = -a*{AB}^n = 0$ since $ω_{AB} = 0$.

y: $a*{B}^n = -a*{A} + a_{AB}^t$

From the second equation and $a*{AB}^t = α*{AB} * AB $

we can find that $α_{AB} = 12$

(because I found that $a_{B}^n = 8$).

At that point I don't know what to do. Am I doing it right and how can I find the acceleration of B and instant centre of acceleration? (using known formula with $tan$ etc.)