Space Curvilinear Motion question. Very hard

[akufrd]

1. The problem statement, all variables and given/known data

Diagram from picture attached
R= 200 + 50sin (2pi nt), n=2
R= 200+ 50sin(4pi*t) mm
ThetaDot = 120 rev/minute
= 12.566 rad/s
Gamma(Y) = 30 degrees

Calculate acceleration when velocity is max

2. Relevant equations
ar= rDotdot - r*(thetaDot)^2
aTheta= r*thetaDotdot + 2*rDot*thetaDot
az = zDotdot

3. The attempt at a solution

At max velocity, cos(4pi*t) = 1
so t=0, 0.25, or 0.5
so sin(4pi*t) will always = 0

Resolve R to get the radius at max velocity
r = Rsin(theta)
= (0.2+ 0.05sin(4pi*t)) * sin 30 , sin(4pi*t) = 0
=0.2 * sin 30 = 0.1 m

Differentiate R and we will get oscillation velocity
dR/dt= 0.2pi*cos(4pi*t)


Resolve the velocity to get Vr
Vr= [0.2pi*cos(4pi*t)] * sin 30
=0.1pi*cos(4pi*t)
= 0.1pi

Differentiate dR/dt would get the oscillation acceleration

d2R/dt2= -0.8*pi^2*sin(4pi*t)
at max velocity, -0.8*pi^2*sin(4pi*t) = 0

So all there is left is aTheta
aTheta= r*thetaDotdot + 2*rDot*thetaDot
= 0 + 2 * (0.1pi) * 12.566
= 7.895

a of ball = aTheta
=7.895 m/s^2

But the real answer is 17.66!! I asked my friends, my senior, and even my TUTOR cant answer the question! This is just an exercise, not an assignment.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[EddieHimself]

You have to add in the acceleration towards the centre of the circle. That works out at \dot{\theta}2Rwhich is 12.572x0.1= 15.8. Then take the resultant of the 2 components, \sqrt{7.8952+15.82} = 17.66.