Question

Review The spool has a mass of 50 kg and a radius of gyration ko = 0.270 m (Figure 1) Part A If the 23-kg block A is released from rest, determine the velocity of the block when it descends 0.6 m Express your answer to three significant figures and include the appropriate units. μΑ ? VA= Value Units Submit Request Answer Figure < 1 of 1 Provide Feedback Next > 0.3 m 0.2 m 09

Answer 1

Date Page (azm K Spool mal 23 Kg v=o A of the block A MA 23 kg atto mall of the spool = sokg mg = mg = Sokg radius of gyration Ko= 0.27m The block A is released from neste First we draw free body A > T= Tension force in the rope acceleration o of the block when it released ad T = MAXA diagram of block So mag "MAG Free body diagram of spool - Tension force (I) produces Porque about ou TXO. 2 = Ispool 2) Ispool Moment int of inentia of Spool about o. a angular acceleration of Ispool - Mspool & Ko² spool Ispool = 5000 127) Ispool = 3.6 45 kg m 2

a Cangular acceleration) = a= tangential acceleration is equal to acceleration blocka. of Txo.2 3.645 x a O. 2 3.645 a 0.04 from equation value of T is MAJ - MAR So MAG-MAR = 3.645a 0.04 23x9.81-23a 9.4.12sa a = 8.842 m2 1.977m52 So acceleration of blocka) = y. 977m5 Now we calculate velocity of block when it decends 0.6m. According to and initial velocity = 0, a = 1.977mg-2 s (distance) 0.6m v=final velocity claw motion u²+2as 02 + 2 x 1.977X0.6 2.37 248m st of

e = 2.37240 o = √2.37240 v= 7.540 ms-t block when it decends o.6m is So veloc city of 2.540 m/s.