What goes up sometimes stays there

the angle between the line from rest position of coin to the center of the sphere and the line from the spinning coin to the center, k = cos^-1 (.05/.1) = 60 degree. using conical pendulum formula, we have T sin k = ma, and T cos k = mg. Divide the former and the later formula we get tan k = a/g sqrt(3) = a/9.8 a = 16.97m/ss

The coin comes at rest when the net force acting on it is zero. There are three forces acting on the coin: force of gravity, normal force, and force from the acceleration. Let m be the mass of the coin. Then the force of gravity is mg. Since it is given that the coin is at height 0.05m, and the radius of the bowl is 0.1m, then the line of normal force makes a 30 degree angle under a horizontal line. Since gravity and the normal force are the only forces with a y-component, then they cancel out. Hence

Fn = mg/sin30

Similarly, since gravity and the normal force are the only forces with an x-component, then they cancel out. Hence

Fn = ma/cos30

Combining the two equations, we get

a = g cos30/sin30 = 16.974 m/s2

If the coin stays still with respect to you, then the vertical component of the acceleration is zero, while the horizontal component matches your acceleration. The vertical forces on the coin are the force of gravity and the vertical component of the normal force. If $\theta$ is the angle between the vertical axis passing through the center of the hemisphere and the line between the coin and the center of the hemisphere, and N is the magnitude of the normal force, then the net vertical force is $-mg+Ncos \theta=ma_y=0$ . N is therefore $N=mg/cos \theta$ . The horizontal component of the normal force generates the horizontal acceleration $a$ of the coin, and so $ma_x=Nsin \theta = mg tan \theta$ . $cos \theta$ is given by $(0.1-0.05)/0.1=0.5$ and so we can solve for $a_x=16.97~m/s^2$ .