(10%) Problem 2: You have a horizontal grindstone (a disk) that is 89 kg, has a
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(10%) Problem 2: You have a horizontal grindstone (a disk) that is 89 kg, has a 0.36 m radius, is turning at 88 rpm (in the positive direction), and you press a steel axe against the edge with a force of 24 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 of the grindstone in rad/s*. Grade Summary Deductions Potential = 0% 100% sin0 tan() | | ( Submissions Attempts remaining: 2 cos cotanO asin) acosO atanO acotan)sinh(0 cosh0 tanh) cotanh0 Degrees Radians % per attempt) detailed view END CLEAR Submit Hint I give Hints: 1 % deduction per hint. Hints remaining: 3 Feedback: deduction per feedback. - 50% Part (b) What is the number of turns, N, that the stone will make before coming to rest?Explanation / Answer
(A) N = F = 24 N
f = uk N = 4.8 N
Applyng net torque = I alpha
0.36 x 4.8 = (89 0.36^2 / 2) (alpha)
alpha = 0.30 rad/s^2
(B) w0 = 88 rpm = 88 x 2pi rad / 60 s
w0 = 9.22 rad/s
wf^2 - w0^2 =2 alpha theta
0^2 - 9.22^2 = 2(-0.30)(theta)
theta = 141.5 rad
tunrs = 141.5 /2pi = 22.5
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