In the figure, a constant horizontal force of magnitude 47.0 N is applied to a u

Question

In the figure, a constant horizontal force of magnitude 47.0 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 25.9 kg, its radius is 0.288 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?

Explanation / Answer

given

mass of cylinder m = 11kg

applied force F = 12N

radius of cylinder r = 0.288m

from the parallel axes thoerem to the cylinder

Ip = ( 1/2) MR2 + MR2

   the torque is due to the applied force and is ? = Fapplied ( 2R)

where Ip = ( 1/2) 25.9 ( 0.288)2 + 11 ( 0.288)2 = 1.986

               Ip ? = Fapplied ( 2R)

                   ? = Fapp ( 2R)/ Ip

                      = 47 N ( 2*0.288)/1.986 = 13.63 rad/s2

a) magnitude of accleration

                 a= R? = 0.288m( 13.63) rad/s2 = 3.9254m/s2

b) angular accleration ? = 13.63 rad/s2

c) applyinf newtons second law

47 N - f = Ma

there fore f = 47 N - ( 25.9kg) ( 3.9254 m/s2 ) = - 54.66N

the friction force is found to point right ward with magnitude 54.66 N

f = ( 54.66 N) i^

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