WA 121 HW 8 x www.webassign.net/us x Ze Physics for Scientists ar x CPhysics que
ID: 1471477 • Letter: W
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WA 121 HW 8 x www.webassign.net/us x Ze Physics for Scientists ar x CPhysics question l cheg x Physics Q on Error Propa x LP X WebAssign c fi b www.webassign net Responses/last?dep 12715986 t/web/Student/Assignment Myuw W canvass soUNDcLOUD BET c (G Phys 121A Fall 20 sP Phys 121A aut 15 WA WebAssign LoG IN (G CHEM 237B-ANDER P BIOL 220 A (3 unre (o 237A-ANDERSEN M McGraw-Hill Connec 1. 0112 points I Previous Answers Tipler69.P.106 My Notes A billiard ball that is initially at rest is given a sharp blow by a cue stick. The force is horizontal and is applied at a distance h 2R/3 below the centerline, as shown in the figure below The speed of the ball just after the blow is vo and the coefficient of kinetic friction between the ball and the billiard table is Auk. Use the following as necessary: vo, puk, m for the mass and R for the radius of the billiard ball.) a) What is the magnitude of the angular speed of the ba ust after the blow? (Note Write h in terms of R. and assume no tion the instant the ball is struck.) 5v 3h (b) What is the speed of the ball once it begins to roll without slipping? (c) What is the kinetic energy of the ball just after the hit? Submit Answer Save Progress Screenshot taken Click to view 2. G+ -19 points Tipler6 12. P.028 A large gate weighing 150 N is supported by hinges a the top and bottom of the wooden frame, and is further supported by a wire (Assur Copy to clipboard the y direction is upward.) 1:15Explanation / Answer
Apply Newton's second law
Tnet = F av ( h- R) = Icm * alpha = I cm = w/ delt
solving
wo = Fav ( h-R) del t/ I cm = F av( h-R) del t/ ( 2/5) mr^2
from the impulse formula
F av del t = m vo
wo = F av( h-R) m vo/ F av/ ( 2/5) mr^2 = 5 vo ( h-r)/ 2r^2
(b)
Apply Newton's second law to the ball
To = fkR = I cm alpha
Fy = Fn - mg = 0
Fx = - fk = ma
solving
alpha = uk mgR/ I cm = uk mgR/ ( 2/5) mR^2 = 5 uk g/ 2R
from the rotational kinematic equation
w = wo + alpha del t = wo + 5 uk g/ 2R) del t
a = - uk g
v = vo+ a del t
= vo- uk gt
R ( wo+ 5 ukg/ 2R * del t) = vo - uk g del t
del t = 16/21 ( vo/ uk g
v = vo - uk g ( 16/21) ( vo/ uk) g = 5/21 v0
(c) initial kineti energy
Ki = Ktran+ K rot = 1/2 mvo^2 + 1/2 I wo^2
= 1/2 mvo^2 + 1/2 ( 2/5* mR^2 ( 5 vo/ 3R)^2
= 19/18 * mvo^2
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