Respuesta :
Answer:
power required = 1123.07 W
flywheel inertia  = 0.479 kg/m²
Explanation:
given data
diameter = 0.75 inch = 19.05 mm
thickness  = 0.375 inch = 9.525 mm
shear strength = 60,000 psi = 413.68 N/mm²
speed = 1500 rpm Â
drop motor speed = 10%
punched at a maximum rate = 1 per second
solution
we know here drop motor speed = 10%
so speed will be = 1500 × 10%  = 150
so speed remaining will be
N = (1500 + 1350) ÷ 2 Â
N = 1425 rpm
and ω will be
ω = [tex]\frac{2\pi N}{60}[/tex]   .........1
put here value
ω = [tex]\frac{2 \times \pi \times 1425}{60}[/tex] Â
ω = 149.22
and
area will be
Area = π × diameter × thickness   ....................2
Area = π × 19.05 × 9.525
Area =  570.045 mm²
and
maximum shear forced is required is
maximum shear force = area × shear force   ..........3
maximum shear force = 570.045 × 413.68 Â
maximum shear force = 235816.2156 N
and
energy required for punching that is
energy required = average maximum shear force × displace  ..............4
energy required = ( 235816.2156 × 9.525 × [tex]10^{-3}[/tex] )  ÷ 2
energy required = 1123.074727 N-m
so here power required will be
power required = energy required × no of hole per second    .............5
power required =  1123.07 ×  1
power required = 1123.07 W
and
flywheel inertia is express as
Energy  = I × ω²  × Cs   ....................6
here Cs = [tex]\frac{150}{1425}[/tex] Â
Cs = 0.1052
put value in equation 6 we will get
1123.07  = I × (149.22)²  × 0.1052
solve it we get
flywheel inertia  = 0.479 kg/m²