Can anyone help me with this question? Much Appreciated :) Lamar University PHYS

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Can anyone help me with this question? Much Appreciated :)

Lamar University PHYS xNew Talb × 0 (1) Post Malone-Col × www.saplinglearning.com/ibiscms/mod/ibis/view.php?id-4876230 Sapling Learning macmillan learning Jump to. My Assignment Resources # Attempts Score 03/21/2018 11:59 PM 81.1/100 Gradebook Assignment Inform Print CalculatorPeriodic Table Question 3 of 10 Available From: N Map Due Date Sapling Learning Points Possible: 1 A motorized wheel is spinning with a rotational velocity of u 1.09 rad/s. At 0 s, the operator switches the wheel to a higher speed setting. The rotational velocity of the wheel at all subsequent times is given by Grade Category: G 4 Description: Policies: where c 1.5574 and b-697 sa You can check your a You can view solutio give up on any quest You can keep tryingt until you get it right o You lose 5% of the po answer in your quest attempt at that answ At what time, tamax, is the rotational acceleration a maximum? 6 95 Number .273 ms 95 What is the maximum tangential acceleration, a, of a point on the wheel a distance R center? 1.17 m from its Number 10 2 95 a,1.125 m/ s eTextbook What is the magnitude of the total acceleration of this point? OHelp With This Topi Number a2.01739 m/s OWeb Help & Videos Previous Give Up & View Solution Check Answer NextExit Technical Support a Hint © 2011-2018 Sapling Learning, Inc. about us privacy policy terms of use contact us help Type here to search

Explanation / Answer

1)

a = dw/dt = w0* 2bt/(c+bt^2)^2

da/dt =0

c = 3bt^2

tmax = sqrt(c/3b) = sqrt(1.5574/3*6.97)s

tmax = 0.273 s

2)

tangential acceleration at t = 0.273

at = R*dw/dt

at = 1.17*w0* 2b*0.273 / (c + b*0.273^2)^2

at = (1.17*1.09*2*6.97*0.273) / (1.5574 + 6.97*0.273^2)^2

at = 1.125 m/s^2

c)

w = w0*atan(c+b*t^2) = 1.09*atan(1.5574 + 6.97*0.273^2) = 1.223 /s

ac = wR^2 = 1.17^2*1.223 = 1.674 m/s^2

a = sqrt(ac^2 + at^2)

a = sqrt(1.125^2 + 1.674^2)

a = 2.017 m/s^2

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