Question

A fly wheel of mass 12.5kg and diameter 0.36 m rotating at 90rpm has its speed increased to 720rpm in 8s. Find the torque applied to flywheel

Answer 1

Hello dear,

◆ Answer-
τ = 3.34 Nm

◆ Explanation-
# Given-
m = 12.5 kg
r = 0.36/2 = 0.18 m
f1 = 90 rpm = 1.5 rps
f2 = 720 rpm = 12 rps
t = 8 s

# Solution-
Rotational acceleration is-
α = (ω2-ω1) / t
α = 2π(f2-f1) / t
α = 2π(12-1.5) / 8
α = 8.247 rad/s^2

Moment of inertia is -
I = mr^2
I = 12.5 × 0.18^2
I = 0.405 kgm^2

Rotational torque is calculated by-
τ = Iα
τ = 0.405 × 8.247
τ = 3.34 Nm

Therefore, torque required is 3.34 Nm.

Hope this helps you...

Answer 2

Answer:

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