Wiley PLUS: My WlleYPLUS Halliday, Fundamentals of Physics, 10e GENERAL PHYS US

Question

Wiley PLUS: My WlleYPLUS Halliday, Fundamentals of Physics, 10e GENERAL PHYS US CPH 111/112/113) Assignment Gradebook. ORION nment FULL SCREEN PRINTER VERSION Chapter 10, Problem 006 The angular position of a point on the rim of a rotating wheel is given by e- 4.27t -1.47B 2.82t, where e is in radians and t is in seconds. What are the angular the at (a) 1.98 s and (b) t 9.97 s? (c) What is the average acceleration for the time interval that begins at 1.98 sand ends at t 9.97 s? What are instantaneous angular accelerations at (d) the beginning and the end of this time interval? (a) Number (b) Number (c) Number (d) Number (e) Number Click if you would like to show work for this question: open show work Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER copyright c 2000-2017 by John wary Sons, inc or related companies. Al rights reserved. ocy Polky I 2000 2012 Dohn.wley sons, Inc. All Rights Reserved. A Division of xobn wley s sons Inc.

Explanation / Answer

given theta = 4.27t - 1.47t^2 + 2.82t^3

angular speed w = dtheta/dt

w = 4.27 - 2.94t + 8.46t^2

instantaneous acceleration alpha = dw/dt

alpha = -2.94 + 16.92t

(a)

t = 1.98 s

w = 4.27 - (2.94*1.98) + (8.46*(1.98)^2)

w1 = 31.6 rad/s

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(b)

t = 9.97 s

w2 = 4.27 - (2.94*9.97) + (8.46*9.97^2)

w2 = 816 rad/s

====================

(c)

average angular acceleration alpha = (w2-w1)/(t2-t1)

alpha = (816-31.6)/(9.97-1.98) = 98.2 rad/s^2

======================

(d)

at t = 1.98 s

alpha = -2.94 + (16.92*1.98) = 30.6 rad/s^2

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(e)

alpha = -2.94+(16.92*9.97) = 165.7 rad/s^2

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