In the figure, a constant horizontal force F app of magnitude 57.0 N is applied

Question

In the figure, a constant horizontal force F app of magnitude 57.0 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 11.1 kg, its radius is 0.641 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

When the cylinder rolls, there is a force F=15N at the top and a static frictional force f preventing slipping.
everything in usual notation,
acceleration A=R*a (a=angular acceleration)
Newton's second law implies'
MA=F-f (resultant force)
also for torques,
Ia=FR+fR (for the axis passing through the centre)
FRom the two equatins,
2F=mA+Ia/R
i.e. 2F=maR+Ia/r
a=2F/(mR+I/R)

a=2*57/(11.1*0.641+2.28/0.641) =10.682
I=0.5MR^2= 0.5*11.1 *0.641^2 =2.28
a= 10.682m/s2
therefore, A=aR =10.682*0.641
=6.8472

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