What is the magnitude of the acceleration of cylinder's com?

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A uniform solid cylinder with a mass of 19 kg and a radius of 0.11 m is subjected to an 18 N horizontal force applied via fishing line. The discussion revolves around calculating the acceleration of the cylinder's center of mass, with participants suggesting the use of torque and Newton's equations. The parallel axis theorem is introduced to help understand the moment of inertia in relation to the cylinder's motion. There is confusion regarding the net torque calculation and the forces acting on the cylinder, prompting further clarification on the number of forces producing torque. The conversation emphasizes the importance of accurately applying physics concepts to solve the problem.
JessicaHelena
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Homework Statement


In the figure below, a constant horizontal force
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app of magnitude 18 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 19 kg, its radius is 0.11 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?

Homework Equations

The Attempt at a Solution



I'm confused... I feel like I could use torque — I could set the bottom point as the origin, and then tau_net = 18 x 2 x 0.11 + F_cm x 0.11.
However, I couldn't really get further than that and I am really confused. Could someone help me?
 

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Taking torque about the bottom point is a convenient way of eliminating
the frictional force.
Are you familiar with the parallel axis theorem?
You can also apply Newton's equations about the center of mass
to eliminate the frictional force.
 
@J Hann — what's the parallel axis theorem?
 
JessicaHelena said:
@J Hann — what's the parallel axis theorem?
The moment of inertia I of a body about any axis is equal to the moment of inertia
I CM of the body about a parallel axis through its center of mass plus the mass M of the
body times the square of the perpendicular distance L between the axes:
I = Icm + M L^2.
For a cylinder about an edge I = 1/2 M R^2 + M R^2 = 3/2 M R^2
 
JessicaHelena said:
tau_net = 18 x 2 x 0.11 + F_cm x 0.11.

Taking the torque around the contact point was a very good idea. (Why?) However, your equation for the net torque is wrong. How many forces are there acting on the cylinder? And how many of those produce a torque about the contact point?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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