Numerical:

$m = 5\ kg$

$M = 25\ kg$

$F = 10\ N$

$q = 1$

[A] When, $r_1 = 150\ mm$, $N_1 = ?$

$h_1 = \sqrt{250^2 – 150^2} = 0.2\ m$

$0.2 = h_1 = \frac{895}{N_1^2} [ 1 + \frac{25}{2 \times 5} (1 + 1)]$

$N_1 = 164 \ rpm.$

For $r_2 = 200\ mm$, $N_2 = ?$

$h_2 = \sqrt{ 250^2 – 200^2}$

$h_2 = 0.15\ m$

$h_2 = \frac{895}{N_2^2} ( 1+ \frac{M}{2m} (1 + q))$

$0.15 = \frac{895}{N_2^2} (1 + \frac{25}{2 \times 5} (1 + 1))$

$N_2 = 189.208\ rpm.$

Range = $N_2 – N_1$

$= 25.208\ rpm.$

Sleeve lift = $x = 2 (h_1 – h_2)$

$= 2(0.2\ – \ 0.15)$

$= 100\ mm = 0.1\ m$

Case 1: Neglect friction force:

$C = \frac{N_2 - N_1}{N_1}$ $= \frac{Range \ of \ speed}{Minimum \ speed \ of \ rotation}$

$C = 0.152$

$p = C[mg + Mg]$

$= 0.192 (5 \times 9.81 + 25 \times 9.81)$

$P = 44.7336\ N$

$Power = p \times x$

$= 44.7336 \times 0.1$

$Power = 4.47\ N.m$

Case 2: Friction force of 10N:

1] for $r_1 = 150\ mm, \ N_1 = ?$

$h_1 = 0.2\ m$

$0.2 = \frac{895}{N_1^2} [ 1+ \frac{25 \times 9.81 – 10}{2 \times 5 \times 9.81} (1+1)]$

$N_1 = 161.05\ rpm.$

2] for $r_2 = 200\ mm, \ N_2 = ?$

$h_2 = 0.15\ m$

$\therefore$ $0.15 = \frac{895}{N_2^2} [ 1+ \frac{25 \times 9.81 + 10}{2 \times 5 \times 9.81} \times (1+1)]$

$N_2 = 192.3965 \ rpm$

Range = $N_2 – N_1 = 31.3465 \ rpm.$

$\rightarrow$ sleeve lift $= x = 0.1\ m$

$\rightarrow$ $C = \frac{N_2 – N_1}{N_1} = 0.195$

$\rightarrow P = c (mg + Mg + F)$

$= 0.195 [ 5 \times 9.81 + 25 \times 9.81 + 10]$

$P = 55.3385\ N$

$\rightarrow power = p \times x$

$= 55.3385 \times 0.1$

$Power = 5.53\ N.m$