Question

In the figure, block 1 has mass m - 490 g, block 2 has mass m2 - 5409, and the pulley is on a frictionless horizontal and and has radius R - 4.7 cm. When released from rest, block 2 fois 79 cm in 4.6 s without the cord slipping on the pulley (a) What is the magnitude of the acceleration of the blocks? What are (b) tension Tz (the tension force on the block 2) and (c) tension T, (the tension force on the block 17 (d) what is the magnitude of the pulley's angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g - 9.81 m/s. Units v (a) Number (b) Number (c) Number (d) Number Units Units > > > > Units Units

Answer 1

Sol.a. 12 : 06 m2 g → Acceleration of block 2 y = Yo + Vot + t ext? T, here, yo=0 & Vol AT2 29 : y = ₂ at ² => a= 2x0,79 ma = 0.075 m/s² (4.6)2 block 1 has same acceleration in opposite direction. Ala = 0.075 m/s² cupward) & ab 20.075 m/s2 (clownward) b. From, Newton's and law, for block 2. = m₂ a => mag - Tz = m₂c :. Ta 540x10-3 ( 9.84 -0.075 ) T = 5.26 N Ary i. E FJ - :. To ç For block 1, Efy = M, a => mig-T, = mi(-a) = 0.49 ( 9.84 +0.0 75 ) 4.84 N : [T, = 4.84N) Ans · T

Q= a R こ F1.60 rad 0.047 S2 d. Since the cord does not slip on the pulley, the tangential acceleration of the pulley is equal to the acceleration of the blocks, The cingular acceleration of the pulley is, 0.075 az XR =) :.12= 1.60 rad/s2 Ang e. The rotational inertia of pulley is, Tz1a = Te R - T; A => I = (T2 - Ti) Rae I = (5.26 -4.84) 0.047 I=0.0124 kg.m2) -Ang .. 1.60 ::.