The mass of each ball is $2 \ kg$. if the speeds @ the two extreme positions are $400$ and $420\ r.p.m.$

Find:

1] The initial compression at the central spring.

2] The spring constant.

Initial compression of spring ($\delta) = \frac{S_1}{S}$

Diagram

$N_1 = 400 \ rpm$

$N_2 = 420 \ rpm$

$w_1 = 41.888 \ rad/sec$

$w_2 = 43.98 \ rad/sec$

$Fc_1 = mr_1 w_1^2$

$= 2 \times 0.08 \times 41.888^2$

$FC_1 = 280.736 N$

$FC_2 = mr_2 w_2^2$

$= 2 \times 0.12 \times 42.98^2$

$FC_2 = 404.217 \ N$

For 1st position:

$\sum M_0 = 0 \ \curvearrowright$ +

$(\frac{Mg + S_1}{2}) (Y) – FC_1 (x) = 0$

$\frac{S_1}{2} (Y) = FC_1 (X)$

$S_1 = FC_1 \times 2$

$= 280.736 \times 2$

$S_1 = 561.46 N$

For second position:

$\sum M_o \ \curvearrowright$ +

$(\frac{Mg + s_2}{2}) (Y) – FC_2 (X) = 0$

$\frac{S_2}{2} (Y) = Fc_2 (x)$

$S_2 = 2Fc_2$

$S_2 = 928.434 N$

$h = \frac{Y}{X} (r_2 – r_1)$

$= 120\ – \ 80$

$h = 40\ mm$

Stiffness = $\frac{S_2 – S_1}{h}$

$= \frac{928.434\ – \ 561.46}{40}$

$S = 9.17 \ N/mm$

Initial compression of spring = $\delta = \frac{S_1}{S}$

$= \frac{561.46}{9.17}$

$\delta = 61.199 \ mm$