Question

The disc, with r=72 mm and mass =2.1 kg, turns without slipping with respect to the fixed ring at point C and is attached to the thin bar AB, lenghth=157 mm and mass=1.5 kg, at point B. If the turning radius of the disc is 67 mm and the system starts from rest at an angle θ = 74°, find the angular velocity for the bar AB when the angle θ = 10°.

В А. ө

We need to use the approach of K1 + P1 = K2 + P2, Total energy formula.

Answer 1

Som! 8 B 157 mon hel oz ojc b = lsino = 1578 6674 - 150.92 mm 62 = 15751910 - 27.26 mm Length of so = 15mm mass of AB = 3.54kg at B. Disc radius = 720mm 2.1kg 02=10) Ipisc = mko 2 Irod = (mx ) Using energy conservation beco, woo KI + 1 [w? 2 + {musi I mu? Į TW hod Disc (2.1x9.81 x 150.62 5-335 Iwo Nao, Ri=fi& mo?) 12 P = (mgh Daust migh Darm (2.1X4.81 +(1.5)*9.81X/10-92 K2 = 14 mu?+ ? [w]gout le molts fiablece 4 x2-18027 + 2 2 1 X 0.032 ) 1x0:09]o?? 20-153902 + - xl-5x u? 4x9-SX0-153 + 1o.un 2 XV 0.15 12.1 1.5X0-150 + 2.140.0727 K₂ = 1.955 02

o+ 5.33= 0.9628 + 1.955u? ,حبه لي P2= mighz (1 5+2.1) 89-81*27.26 - iz 1 = 0096275 Using Energy conservation: KHPI = Kat P2 = v=1.5mis (0-10)- Hence, angular velocity of go when los 0.157 Was gossu rad sec ICS Scanned with CamScanners
II was not clear about turning radius and hence i have solved this question by taking rod length as given in the problem.If you get this then solve by following same procedure .

Thank you!!