Question

1. Consider a 10.0 kg solid steel ball with a radius of 8.0 cm (0.080 m) rolling down a track of height h = 4.5 m without slipping. 4.50 m PLEASE SHOW ALL WORK AND CALCULATIONS. Answers W/O derivations will receive no credit. Partial Credit will be rewarded for answers with some errors. a. Calculate the moment of inertia for the ball b. Calculate the initial potential energy of the ball

Answer 1

1) as manent of inertia of ball = 2 2 mr² = 2x10x10.08) kg m² 0.0256 kg m² - by instral potential energy - mgr = (10x9.8x4.50) J 441 J. Kinetic enery Portakiadenergy at - translational rotational the top KE KE mign = mv2 + 1/2 T. w2 = 1 mv²t x 2 / 3 montowa muz [v=rw] mgn = 7 omuz ..Valign X 9:844:50 v = 7.94 mil. tranflatinal KE mv. x 10 x (7.94 = 315.22 J. 7 as moment of inestia of bulley mara - 2x (0.2012 kg ma = 0.08 kg-ma 7 mig-T,= m, a.-(1) (T. -T.) r = IX = mpre xa a Ti-T2 - mpa. -60) m t1! moar it) Adding (i),Gi) and (ii) we get T2 - m29 ema - I mig-m₂ g = a (mitmpt m.) langia on Roth attached mitmptn, to bucket is To 10-4 9.8 = 3.675 mig? a. 10+2 +4 3 mig-m, a = 1019.8-3.675)= 61.25 r. 12. = m2 Sta) 4(9.8+ 3.675) = 53.9ır. - Tenzin on reope attached to lead weight 53.90r. d> : Acceleration is 3.675 mise Net forane on parlling = (T-T2JXP = (61.25-53-9)x 0.20 = 1.47 N-mi