Force on a pin from a pendulum and a string

[magnesium12]

Homework Statement

The simple 2-kg pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at B and continues in the smaller arc in the vertical plane. Calculate the magnitude of the force R supported by the pin at B when the pendulum passes the position θ = 30°.

Relevant Equations

ΔU = ΔT (T = kinetic energy, U = potential energy)

T = 0.5mv^2

U = mgh

∑F = ma

a_n = v^2/r (acceleration in normal direction)

1. Determine the velocity of the ball when it is 30degrees from the horizontal:
U1 = mgh = mg(0.8m)
U2 = mgh = mg(0.4+0.4cos(30)) = mg(.74641)
ΔU = U2 - U1 = mg(.74641 - .8) = mg(-0.051433)

T1 =0
T2 = 0.5mv^2
ΔT = T2 - T1 = 0.5mv^2

ΔU = ΔT
mg(-0.051433) = 0.5mv^2 ====> v = 1.025394

2. Use the velocity to find the acceleration in the normal direction (pointing along the string towards the pin):
a_n = v^2/r = 1.025394^2/(0.4m) = 2.62858

3. Sum up the forces along the normal direction acting on the ball to find the tension (S) of the rope segment from the ball to pin:
∑F = S - mgcos30 = ma_n
S = ma_n + mgcos30 = (2*2.62858) + (2*-9.81*cos(30)) = 5.257163 - 16.991418 = -11.6809
S = -11.6809

4.Sum up the forces acting on the pin using the xy coordinate system (?)
This is where I'm confused. The sum of forces on the string should be zero, because the pin isn't moving, shouldn't it? So:
∑F_y = 0 = S_y + R (R = tension in the vertical string segment from pin to ceiling)
So R = S_y = Scos30 = -11.6809cos30 = -10.1159

The answer should be 46.7N but I don't understand how to get there. Can someone please point out where I'm going wrong?

 

[BvU]

Very early on: in $\Delta U$

 

[magnesium12]

Very early on: in $\Delta U$

Should I be breaking up the motion to account for the shorter string length?
So from top to vertical (total bottom) it would be
ΔU =ΔT
mg(0.8) = 0.5mv_1^2

Then do the same analysis from bottom to the height at 30deg:
U1 = 0
U2 = mg(0.4-0.4cos30)
T1 = 0.5mv_1^2
T2 = 0.5mv_2^2
ΔU =ΔT
mg(0.4-0.4cos30) = 0.5mv_2^2 - 0.5mv_1^2

Am I thinking about this the right way?

 

[BvU]

Should I be breaking up the motion to account for the shorter string length?

Either that, or establish a correct height variable: your original $\Delta U$ for $\theta=0$ would come out as zero !

Then:

The sum of forces on the string should be zero, because the pin isn't moving

is correct, though the logic looks awful. And the pin is held in place by the wall.

 

[haruspex]

Should I be breaking up the motion to account for the shorter string length?

No, you just need to be consistent in your choice of origin for vertical heights and which way is positive. Your U1 is the PE at A above an origin at the base of the swing, while U2 is the PE at A above an origin at the final point.
(It is usually worthwhile checking that a numeric result looks sensible. Did it not strike you as unlikely that the net fall in height was so small?)