In the figure, a constant horizontal force of magnitude 79.5 N is applied to a u
ID: 1373609 • Letter: I
Question
In the figure, a constant horizontal force of magnitude 79.5 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 20.3 kg, its radius is 0.259 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 = 20.3 kg
applied force F = 79.5 N
radius of cylinder r = 0.259 m
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) 20.3 ( 0.259)2 + 20.3 ( 0.259)2 = 2.043
Ip ? = Fapplied ( 2R)
? = Fapp ( 2R)/ Ip
= 79.5 N ( 0.518)/2.043 = 20.157 rad/s2
a) magnitude of accleration
a= R? = 0.259 m *20.157 rad/s2 = 5.22 m/s2
b) angular accleration ? = 20.157 rad/s2
c) applyinf newtons second law
79.7 N - f = Ma
there fore f = 79.5 N - ( 20.3 kg) ( 5.22 m/s2 ) = - 26.466 N
the friction force is found to point right ward with magnitude 26.466 N
f = ( 26.466 N) i^
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