Question

Question Certi Status The pendulum consists of a 6.kg slender rod fixed to a 15 kg disk. If the spring has an unstretched length of 0.2 m. determine the angular velocity of the pendulum when it is released from rest and rotates clockwise 90° from the position shown. The roller at allows the spring to always remain vertical C 0.5 m k = 200 N/m B A D 0.5 m – 0.4 m Toom Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers

Answer 1

Date: Page no Solna Kinetic energy: The mau moment of Inertia of the Pendulum about B is, Te [24][0-9]+660-458] + [3 (15)(021 +15(4013] To = 20.07 kg. m² Tz- I Ig W² Ź (20:07) 62 Thus, T = 10.035 W² Since the Pendulum is released from rest, T=0 0.5m 02 m 0.4 m 10-5m Datum E MW 2 (yd),

Page no Potential Energy - with refrence to the datum sets (above figute), the gravitational Potential energies of the Pendulum when it is at Position ③ & ④ are, (Vg), - Me g (Y2), + mag (Yd); = 0 (Vg) = Me g(Y&)2 + mag (yd). (Vg), = 6(9:81)(-0.5) + 15 (9-81) (-11) = -198.295 J The stretch of the spring when the Pendulum is at Positions and ② are, x=0.5-0.2 -0.3 m I-0.2 -0.8 m Thus, the initial and final elastic Potential nezgils of the spring are, (vel IK202 (200) (0:33 9.00 J - 2 2 (Ve)2 = I KX ² = † (200) (0.82) - 64.0J 2 2 Conservation of Energy :- T + ₂ = Tz + V2 0 + (0 +9:00) = 10:035 W2 + (-191295 +64.0). w = 3:6854 tad /s (OR) W =3.685 tad's Ans.
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