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diamong [38]
3 years ago
10

A constant torque of 29.1 N · m is applied to a grindstone whose moment of inertia is 0.142 kg · m2 . Using energy principles, a

nd neglecting friction, find the angular speed after the grindstone has made 14.1 rev, assuming it started from rest. Answer in units of rev/s.
Physics
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

76.0198 rev/sec

Explanation:

Given

Torque ( T ) = 29.1 N

Moment of inertia = 0.142 kg.m²

Ф = 14.1 rev

F =  \frac{1}{2} I w^{2}

w = angular speed

F = T ΔФ

w² =( 2TΔФ)/I    = \frac{2 *29.1*14.1}{0.142}  

w = √5779.01    = 76.0198 rev/sec

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\displaystyle \vec{d}=

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