Question

Please include all steps/formulas and Diagrams (Fbds)

The 20-kg square plate is pinned to the 5-kg smooth collar. Determine the initial angular acceleration of the plate when P = 100 N is applied to the collar. The plate is originally at rest. (Answer: 35.4 rad/s?) P-100 N 300 mm 300 mm

Answer 1

Here given ? Msp = 20kg . Misc = 5kg. Р =100N L 300 mm 0.300 m ny SC R Ch.0-300m- t COG I coule Now, we honee vcowrider the forces was equilibuuim; Hence, the total forces equilibrium valong a-duection; (Fxd sp P + (Fr)sc equilibine cum valong y- But, for the total forces on direction j (Fy sp O. rabout A; Now, taking the moment (F) asp R Icoex«. C

know Force , f = Mass & Acceleratius Snce, we FEMA and © becomes; > then, the equations ② 0 (Msp Agg) - P +(Msc Asedy (D) (Mapa spally C-> Msp Aspot R Icos xүр Since the IcoG becomes ; The rotational moment of Inertia, mero Emce, M = Pbd dumped mars) lehere, I COG - e (balss + db) M (62+d²) I cosi 12 Here badah = 0.300m ; M = M. = IcoG Mspx ( 2² + 4² ) 12 ICOG 20kg *60.300m + 0.300m) 12 IcoG= 0.3 m 4

Now, substituting value of Icon im equatins Msp x Aspg XR = 0.3m4x casp. R 2=0.300m On subslitutis ; N' Cos45= R xL los 45º = 0.3m4x asp. 20kg x Aspg? 20kg x Aspqx 0.300 mx Cos 45° 0.3m 4 no exasp. asp = 14.142 Aspq Ar = 0. 0509 «гр Then, we honce a relations for the caccelerator at porst A. of the smooth plate; Rz Aspq asp. Cosq5 = R on substitution) Llos 45º = Asc 0.0707 asp asp. 3) 0.3 los 45° - Asc 010707 asp asp

=) 0.300545° asp Asc - 0.0707 dsp. Asc = 0.2828 asp Now, the equation h becomes, Msp Aspg - p + (Msc Ascg = 0 Qo68 x 0.09 09sp - 100N+ 56g 10:28 21) asp = 0. 1.414 45p - 100 + 1.4146sp = 0. =) Xsp = 35. 360 rad 1 seconds? Hence, the initial cangular cacceleration of the asp 35.360 sadlan plate , seconds 2