Question

constant torque 2.5 kW motor drives a riveting machine. The mass of the moving parts including the flywheel is 125 kg at a radius of gyration of 700 mm. One riveting operation absorbs 1000 J of energy and takes one second. Speed of the flywheel is 240 rpm before riveting. (a) Determine the reduction in speed after one riveting operation. (b) Modify the mass moment of inertia to the flywheel if the reduction in speed is to be limited to 35 rpm.​

Answer 1

speed after one riveting operation = 230.5 rpm

mass moment of inertia = 9.90 kg

Step-by-step explanation:

given data

torque energy = 2.5 kW = 2500 W

mass = 125 kg

radius = 700 mm

energy = 1000 J

Speed of the flywheel N = 240 rpm

solution

we know here Speed of the flywheel N so here angular speed ω

ω =
`(2\pi N)/(60)` .....................1

ω =
`(2\pi 240)/(60)`

ω = 25.13 rad/s

so here

change in energy is

ΔE = E1 - E2 ..............2

ΔE = 2500 - 1000

ΔE = 1500 J/sec

and

I = mr² .........3

I = 125 × 0.7²

I = 61.25 kg-m²

and ΔE is express as here

ΔE = 0.5 × I × (ω² - ω1² ) ........4

put here value and we get

1500 = 0.5 × 61.25 × (25.13² - ω1² )

ω1 = 24.13 rad /s

and

reduction in speed after one riveting operation will be

N =
`(24.13* 60 )/(2\pi )`

speed after one riveting operation = 230.5 rpm

and

for 35 rpm

ω1 =
`(2\pi 35)/(60)`

ω1 = 3.65

so ΔE will be here

ΔE = 0.5 × mr² × (ω² - ω1² ) ....................5

put here value and we get m

1500 = 0.5 × m (0.7)² × (25.13² - 3.65² )

solve it we get

mass moment of inertia = 9.90 kg