Question

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, and 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.

Answer 1

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