Direction of frictional force on a particle on a plate in circular motion If we have a particle on the surface of a round plate ,at a distance $r$ from the center, that is rotating with an angular acceleration $k$ , and this particle is in equilibrium with respect to the non-inertial system of the plate: What is the direction of the friction between the plate and the particle ? I think that the forces acting on the particle are:


*

*Centripetal force $F_1=m\omega^2 r$      ($\omega$=angular velocity);

*Force which causes the angular accelaration: $ F_2=mkr $ .

 A: It is well known that in polar coordinates $(r,\theta)$ we have
$$\ddot{\vec r} = \left(\ddot r - r \dot \theta^2\right) \hat r + \left(r \ddot \theta + 2 \dot r \dot \theta\right) \hat \theta.$$
This reduces as follows since $r$ is constant ($\dot r = \ddot r = 0$).
$$\ddot{\vec r} = -r \dot \theta^2 \hat r + r \ddot \theta\hat \theta$$
Since the friction force $\mathbf F_f$ is the only forece acting on the object (neglecting forces normal to the plate), we have $\mathbf F_\text{net} = \mathbf F_f$. Substituting your notation $\dot \theta = \omega$ and $\ddot \theta = k$, and then multiplying both sides by $m$, we have the frictional force.
$$\mathbf F_f = \mathbf F_{\text{net}} = m\ddot{\vec r} = -mr \omega^2 \hat r + m r k \hat \theta$$
The right-hand side is exactly what you wrote in the original post. So, your force analysis is correct. 
However, what I am not sure you understand completely is this: the only force acting on the object in the plane of the plate is friction. Therefore, friction must account for BOTH of these components (both radial and angular). The frictional force is the sum of these two and acts neither strictly in the radial or angular directions.
Given this insight, see if you can figure out the direction of the force, since that's what you're really after here.
A: According to me..(there is a possibility that i may be wrong)...there are two directions in which friction is acting...one is in the radial direction and the other is in the tangential direction..the values of friction are m(kr) in tangential direction and mrw^2 in the radial direction (considering nmg greater than m(kr) and mrw^2 where n is coefficient of friction)..since both directions are perpendicular to each other the answer of friction should be √(m(kr))^2 + (mrw^2)^2. And direction is towards angular bisector of radial (towards centre) and tangential (opp. Motion of body) directions.
