Question

In the figure, two 5.10 kg blocks are connected by a massless string over a pulley of radius 2.50 cm and rotational inertia 7.40 x 10-4 kg.m2. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block; the pulley's axis is frictionless. When this system is released from rest, the pulley turns through 0.600 rad in 109 ms and the acceleration of the blocks is constant. What are (a) the magnitude of the pulley's angular acceleration, (b) the magnitude of either block's acceleration, (c) string tension T2, and (d) string tension Tz? Assume free-fall acceleration to be equal to 9.81 m/s2 Ti (a) Number Units < (b) Number Units (c) Number Units (d) Number Units

Answer 1

-4 IM Given m, = M₂ = 5. 10K§ R = Radius of Peelley - 2.500m, 0.025 m 2 Is Rotational inertia = 7.40 x 10 kom Prelley tecoens through 0.600 read in 109 ms. AO = 0.600 radians 109 ms - 109 x10 S 0.109 s enitial angrilare velocity - O tad Using equation fore dotational Kinematics 2 + Wet t 2 2 -3 t = at? Tz out tt + 224² - O - Oy - Wet that? Ot 1 x 2 x (0.109 Δρ. 2 O.GOO M (FBD-M2) 0.600 5,9405 x 10 101.001 pad 52 T 101 Bad Angular acceleration of Pelleg. Using relation between lineare acceleration and angular acceleration. a = RL] pad (0.025 m) x (101 MIS (FBD- BD-M) Tz RT > a 2) T 2.525 m2 + block's acceleration. (FBD-Pzelley) T2 - Ff Applying Newton's Second law to (FBO-M2) Ma ê Applying Newton's second law to FBD-M, M, 8-T, = m, a = Mig-M, a Ti = M, C8-a) Prelley Det torque =( I ->83 -01- 5,- 59 - 12 - 3,- TIR = Toreque (3)-1 & 3 = TR= Toreque (ro)-2 TR - TQ R = 12 MR² x TIR - TR = 12 (T, - T2 R = 22 MR xa. Ti - T2 = 2 ma for -> -> 4 R

Substituting values values in E T = M, Cg-a) » T (5.SOKS) x ( 9.81 m - 2.525 m) 7) T, = 5.10 kg x 7.285 mys? 37. 1535 N - 37.15N - 37. 15 N - T, a