Question

Constants Part A A high-speed flywheel in a motor is spinning at 500 rpm (revolutions per minute) when a power failure suddenly occurs. The flywheel has mass 39.0 kg and diameter 79.0 cm. The power is off for 25.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 150 complete revolutions. At what rate is the flywheel spinning when the power comes back on? IVO AXO ? You may want to review (Pages 278-280) For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Rotation with constant angular acceleretion. - 1.17 rad/s Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Part B How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back IVO AE ? t = S Submit Request Answer Part C How many revolutions would the wheel have made during this time? IVO AEO ? N = геу Submit Request Answer

Answer 1

79cm f 500 rpm when power is on Mass Hy why = 39.0kg. Radius Rayn 39,5 cm 25 79cm 2 The power cut is odd for seconds, and it makes 150 sercultons during this time, using kmematical eqn for ootafsonal motion, ( Winkel + Whoud) to when power cuf is off 150 xv ( 500 rpm + Wannel) 25s. Opiny innof 2 2 Iso ru S 25 500 opm & Whoul, 300 ru 25 sce Winal, 300 tu 500 opm 25 sec Go see/min wonal, 720 opm 500 rpm wanol 220 opm 220 x 20 60 w finul 23.04 rods.

Here we have another Kinematical eqh, Whinul winihal + x,t. winnal - + Lot we struck here, since er don't know time, 500 din 8.3 rev rev/s Now, winney = 500 ser 6os min Now, w final winnot from 2 t wf = 220rpm wf 2:20 3.67 zer/s 50 X = 3.67 – 8,3 25 s -0,1852 rev/e2 8 Now, ean t 8.3 ser + (-0.1852 rena 44.816 seconds 8.3 T 0, 1852 t wehind two final we know, 2 Now, 8.3 ren eris et x 44,816 2 6 a 186 av.