Space Curvilinear Motion question. Very hard

[akufrd]

Homework Statement

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

Homework Equations

a_r= r_Dotdot - r*(theta_Dot)^2
a_Theta= r*theta_Dotdot + 2*r_Dot*theta_Dot
a_z = z_Dotdot

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 V_r
V_r= [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 a_Theta
a_Theta= r*theta_Dotdot + 2*r_Dot*theta_Dot
= 0 + 2 * (0.1pi) * 12.566
= 7.895

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

But the real answer is 17.66! I asked my friends, my senior, and even my TUTOR can't answer the question! This is just an exercise, not an assignment.

 

[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, [itex]\sqrt{7.895²+15.8²}[/itex] = 17.66.