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Unformatted text preview: Chapter 3
17. Since the motor is turned oﬀ, its torque is T = 0, and the only torque acting on the ﬂywheel √ is the retarding one, −5 ω . Then Newton’s second law for rotational motion becomes I √ dω = −5 ω dt with ω (0) = ω0 = 225 (rad/sec) and I = 50 kg/m2 . The general solution to this separable equation is ω (t) = − Using the initial condition, we ﬁnd ω (0) = −0.05 · 0 + C Thus 1 15 − ω (t) = 20 15 − ω (t) . 0.05 when the ﬂywheel stops rotating we have ω (tstop ) = 0 and so t= tstop = 20(15 − √ 0) = 300 (sec). ⇒ C= ω (0) = √ 225 = 15. 5 t + C = −0.05t + C. 2I At the moment t = tstop 19. There are three forces acting on the object: F1 , the force due to gravity, F2 , the air resistance force, and F3 , the friction force. Using Figure 3.11 (with 30◦ replaced by 45◦ ), we obtain √ F1 = mg sin 45◦ = mg 2/2 , √ F3 = −µN = −µmg cos 45◦ = −µmg 2/2 , and so the equation describing the motion is √ √ mg 2 µmg 2 dv = − − 3v m dt 2 2 with the initial condition v (0) = 0. Solving yields √ v (t) = 9.5g 2 + Ce−t/20 ; √ 0 = v (0) = 9.5g 2 + C 132 √ C = −9.5g 2 F2 = −3v, ⇒ √ dv v = 0.475g 2 − dt 20 ⇒ ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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