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ID: 1511035 • Letter: D

Question

D Solution Manual Wiley.pcx e WileyPLUS -> C D edugen.wileyplus.com/edugen/student/mainfr.un. :: Apps HP Connected Watch NHL Live Stre- Watch NBA Live Strea × Physics question Chegg. x ABP Practice Assignment GradebookORION ssignment FULL SCREEN PRINTER VERSION BACK ES Chapter 11, Problem 071 16 41 In the figure, a constant horizontal force app of magnitude 59.3 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 22.3 kg, its 28 067 046 radius is 0.616 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder? 41 63 app 071 Fishing line dy (a) Number (b) Number (c) Number Units Units Units SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM I'm Cortana. Ask me anything 11:54 PM 4/27/2016

Explanation / Answer

Here ,

Fapp = 59.3 N

mass , m = 22.3 Kg

radius , r = 0.616 m

a) let the frictional force is f

acceleraiton of the centre of mass is a

Using second law of motion

Fapp - f = 22.3 * a

59.3 - f = 22.3 * a ----(1)

for the torque

F * r + f * r = 0.5 * m * r^2 * (a/r)

59.3 + f = 0.5 * 22.3 * a ----(2)

sovling 1 and 2

a = 3.45 m/s^2

the magnitude of acceleration is 3.45 m/s^2

b)

angular acceleration = a/r

angular acceleration = 3.45/.616 rad/s^2

angular acceleration = 5.76 rad/s^2

c)

fictional force = 59.3 - 22.3 * 3.45

f = -17.6 N

the frictional force is -17.6 N