Physics
posted by dallas .
A constant torque of 22.9 N·m is applied to a grindstone whose moment of inertia is 0.168 kg·m2. Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 17.1 rev, assuming it started
from rest.
Answer in units of rev/s

M=Iε
ε=M/I = 22.9/0.168 = 136.3 rad/s²
2πN=2πn₀t + εt²/2
2πn=2πn₀ + εt
n₀=0 =>
2πN= εt²/2
2πn= εt => t=2πn/ε
N= πn²/ε
n=sqrt {εN/π}=
=sqrt{136.3•17.1/3.14}=27.24 rev/s
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