1. A uniform spherical globe (1.- Sm,) having a radius of 15 cm and a mass of 0.
ID: 2031209 • Letter: 1
Question
1. A uniform spherical globe (1.- Sm,) having a radius of 15 cm and a mass of 0.05 kg, rotates about its vertical axis on frictionless bearings. A light chord is wound around the equator of the globe, then run over a frictionless pulley (-mand attached to a metal weight as shown in the figure above. The pulley is a solid wheel having a mass of 0.03 kg and a radius of 5 cm. The metal weight has a mass of 0.1 kg. If released from rest, Let us determine the velocity of the metal weight after it has descended a distance of 0.2 m, using conservation of energy of the metal weight if it is a) Write an expression for the gravitational potential energy, U hanging from a hcight of 0.2m. b) Write an expression for the translational (linear) kinetic energy, K, of the metal weight as it is descending. c) Write an expression for the rotational kinetic energy, Kp of the pulley d) Write an expression of the rotational kinetic energy, Ko, of the spinning globe. c) Equate the gravitational potential encrgy,Ug, of the metal weight with the three kinetic terms. PHYSI 11 Benton Spring 2018 OSU Physics Dept. f Substitute in the appropriate expressions for the moments of inertia of the pulley and the globe and substitue in appropriate values of the angular velocities of the pulley and the globe in terms of tangential velocity and radius. Then simplify the algebra. g) Rewrite the equation to solve for velocity, plug in values, and determine the desired velocity of the descending weightExplanation / Answer
a) Ug = mgh = 0.1*9.8*0.2 = 0.196 J
B) Kt = mV2/2 = 0.1*V2/2 = 0.05V2 J
c) Kp = Ipwp2/2 = mprp2wp2/4
= 0.03*0.052wp2/4
= 1.875x10-5wp2
d) Kg = Igwg2/2 = mgrg2wg2/3
= 0.05*0.152wg2/3
= 3.75x10-4wg2
e) 0.196 = 0.05V2 + 1.875x10-5wp2+ 3.75x10-4wg2 .....1
f) wp = Vrp = 0.05 V
wg = Vrg = 0.15 V
plugging these in equation 1 we get
0.196 = 0.05V2 + 1.875x10-5 *(0.05 V)2+ 3.75x10-4 * (0.15 V)2
g) V = 1.98 m/s
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