Online HW 08 Begin Date: 10/28/2017 5:00:00 PM-Due Date: 11/10/2017 5:00:00 PM E
ID: 1774277 • Letter: O
Question
Online HW 08 Begin Date: 10/28/2017 5:00:00 PM-Due Date: 11/10/2017 5:00:00 PM End Date: 11/10/2017 5:00 (10%) Problem 6: Suppose you a solid disk) exert a force of 150 N tangential to the outer edge of a 1.27-m radius 72 5-kg grindstone (which is a torque r, in newton meters, you are exerting on the grindstone, relative to the center of mass of the grindstone? Deductions 0% Potential 100% sin0 cotanO asino acos atan0 acotanOsinh0 7 8 9 Attempts remaining: 2 12 3 33% Part (b) what is the angular a celeration of the grindstone 33% Part (c) what is the angular acceleration of the grindstone af in radians per square second, if there is an opposing frictional force of o in radians per square so 22 N exerted 0.35 m from the axis? assuming negligible friction?Explanation / Answer
(i) torque = F * r since the force is perpendicular to the radius
= 150N * 1.27m = 190.5 N·m
(ii) also = I * where I is the moment of inertia
For a uniform disk, I = ½mr² = ½ * 72.5kg * (1.27m)² = 58.46 kg·m²
so
190.5 N·m = 58.46kg·m² *
= 3.25 rad/s²
(iii) Now net torque ' = 190.5N·m - 22N * 0.35m = 182.8 N·m, and so
182.8 N·m = 58.46 kg·m² * '
' = 3.13 rad/s²
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