lp HW 10 Begin Date: 10/16 2017 11:00:00 AM-Due Date: 10/30 2017 11:00:00 AM End
ID: 1790928 • Letter: L
Question
lp HW 10 Begin Date: 10/16 2017 11:00:00 AM-Due Date: 10/30 2017 11:00:00 AM End Date: 10/30/2017 11:00:00 AM (1096) Problem 2: You have a horizontal grindstone (a disk) that is 86 kg, has a 0.34 m radius, is turning at 88 rpm (in the positive direction), and you press a steel axe against the edge with a force of 18 N in the radial direction. ©theexpertta.com 50% Part (a) Assuming that the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration ofthe grindstone in rad/s2 × 50% Part (b) N= 187.21 What is the number of turns, N, that the stone will make before coming to rest? Grade Summary Deductions 10% 90% 7 8 9 HOME Attempts remaining: 3 (5% per attempt) detailed view cotanas cosh0 tanh0 END 50% Degrees C) Radians NO BACKSPACE 5% Submit Hint l grte up Hints: 1 % deduction per hint. Hints remaining: 3 Feedback: 1% deduction per feedback.Explanation / Answer
Given m = 86 kg
radius r = 0.34 m
initial angular velocity omega1 = 88 rpm = 88 * 2pi / 60
omega1 = 9.21 rad/s
normal force N = 18 N
a)
Th firction force on the gridestone is f = mue * N
f = 0.2 * 18
f = - 3.6 N
this force causes a torque which reduces the speed so
torque T = I * alpha
- f * R = I * alpha
- f * R = (m * R^2 / 2) * alpha
- 3.6 * 0.34 = (86 * 0.34^2 / 2) * alpha
therefore alpha = - 0.246 rad/s^2--------------------------------------------Answer
b)
final angular velocity omega2 = 0
from the equation
(omega2)^2 - (omega1)^2 = 2 * alpha * theta
0^2 - (9.21)^2 = 2 * (- 0.246) * theta
theta = 172.406 rad
number of turns N = theta / 2pi
N = 172.406 / 2pi
N = 27.44 turns----------------------------Answer
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