Question

A sphere of radius R can rotate about a vertical axis on frictionless bearings (see figure below). Let the rotational inertia of the sphere be A massless cord passes around the equator of the sphere, over a pulley with rotational inertial and radius r, and is attached to a small object of mass m. There is no friction on the pulley's axle and the cord does not slip on the pulley. At 1 = 0, the mass m has speed v. The system is subject to the regular force of gravity. (a) Using conservation of energy, derive an expression for the speed of mass m after it has fallen for a distance h. Your expression should depend on 1 and 1 (b) For the 3 cases below, evaluate the speed of the falling object in m/s with the following parameters: v = 0, m = 3.0 kg, h = 1.0 m. • Mass of “sphere”: M. = 5.5 kg, radius of “sphere” R= 0.20 m • Mass of pulley: M... = 1.5 kg. Case 1: a. Solid sphere of uniform density. b. Pulley is a solid disk of uniform density with radius r = 0.10 m. Case 2: a. "Sphere" is a thin spherical shell. c. Pulley is a solid disk of uniform density with radius r = 0.10 m. Case 3: a. Solid sphere of uniform density. b. Pulley is an annular cylinder (or ring, see figure below) of uniform density, with radii R = 0.10 m and R = 0.09 m.

Answer 1

This part of the question is based on consercation of energy. The decrease in gravitational potential energy of the block will convert nito kmetic energy of system, mga = KE ofblock + Rotational k. Eof Cophere & Pully - Kinetic energy of block – muz Rotataul Kinectic energy. & Topherus to Ibally wa = Lobhere Re Fully Relation ben je w u= Rw= = R dipwa in vui Splore mgh=km (martty Isphere we are the balley le U agam Ano mtIobkere + Ilcilly Riphere Rot here=55kg @ Now critti al speed of black vi so CR=0.20m; ma bizim) Candi will be Let Esphere = KIMR2 ZK, accorching to Object 3kg, en I pully = ke mpully

v= anga mikim opherpe BZ + Kampulleyna Lingle mq tkim sphere tkampelley. Now for Cape -I- for of there K1= Z for disc k2=1/2 - Us 2x 3x9.8 3+3x55+2x15 SOO = 3.14 ms 3+2:2+0.75 Cape -2 KLAASSA ki foroßhrical shell = 2/3 k2 for disc = y2 =2.02mio U- 3+2455+ ₂x15 Cape-37 ki for ekkere = 2/3 Ks for sing memement of mestial for annullar cylinder =) I= MC R2 + R22) 500 2 NOW US 2 44 ange m+ 3 mophere + Mais CR12 + R22) R,2 50.0 3+227075C1+692) 2.924 M/S

= Cape - I speed =3'14 mio. I solid sphere=2mp2 Cape I speed =2.82 mp Ispherical shell м2- moment of Interial Increase speed deereana ap bighi moment of Incotia Conoume bige Kmectio energy Cape, I ape-UT beed - 3.14 mp3 Idioc=MR,2 ARI speed=2.924 m/s Iannuler dioc ! (R, 2 + R22) as moment of hestia increases speed 2 deerean es