Remaining in the same spot on a spinning disk

[haruspex]

the childs weight creates a torque of R/2*30kg * sin(b)

That is dimensionally wrong. You forgot something.
But please, let us do it the safe way, with free body diagrams, ΣF=ma etc.

 

[Fibo112]

That is dimensionally wrong. You forgot something.

yes I forgot g

 

[haruspex]

yes I forgot g

Right... and please see my edit above.

 

[Fibo112]

Ok. So since the child remains motionless in an Inertial system we have a=0. The forces on the child are the Gravitational force, the normal force of the disc and the friction force of the disc. The component of the Gravitational force which is perpendicular to the disk will cancel out with the discs normal force. The remaining forces on the child are 30kg*g*sin(b) for the remaining gravitational force and a frictional force of equal magnitude and opposite direction to compensate the gravitational force. By Newtons third law the child enacts an equal and opposite force on the disk. The normal component causes no torque. The frictional component causes a torque of T= R/2 30kg*g*sin(b). This causes an angular acceleration of T/Moment of Intertia. The acceleration of the disk at the point of the child is R/2*T/Moment of inertia.?

 

[haruspex]

Ok. So since the child remains motionless in an Inertial system we have a=0. The forces on the child are the Gravitational force, the normal force of the disc and the friction force of the disc. The component of the Gravitational force which is perpendicular to the disk will cancel out with the discs normal force. The remaining forces on the child are 30kg*g*sin(b) for the remaining gravitational force and a frictional force of equal magnitude and opposite direction to compensate the gravitational force. By Newtons third law the child enacts an equal and opposite force on the disk. The normal component causes no torque. The frictional component causes a torque of T= R/2 30kg*g*sin(b). This causes an angular acceleration of T/Moment of Intertia. The acceleration of the disk at the point of the child is R/2*T/Moment of inertia.?

Yes.
Note that this differs from what your equation in post #27 would have given, even after restoring the g factor.

Out of interest, note that your analysis interprets this:

positive y-direction with constant acceleration a relative to the disk

as meaning relative to the tangential acceleration of the disc, not an acceleration of magnitude a relative to the reference frame of the disc. I think you have the right reading.

 

[Fibo112]

I am still bothered by this questions notion of acceleration, it just doesn't make any sense to me to be talking about acceleration without a clear frame of reference as to where this acceleration is taking place...

 

[haruspex]

I am still bothered by this questions notion of acceleration, it just doesn't make any sense to me to be talking about acceleration without a clear frame of reference as to where this acceleration is taking place...

The question states that the child is to stay in the same place in the laboratory frame. That is enough to determine the force the child must exert on the disc, and the disc's consequent acceleration.
The slightly awkward part is that it asks for "the child's acceleration relative to the disk." I now consider that ambiguous. I think what they want is relative to the tangential acceleration of the disc. But if it means relative to the reference frame of the rotating disc then you would have to add the apparent radial acceleration, $-\omega^2\frac R2\hat r$.

 

[Fibo112]

Ok. Thanks a lot for your help.