How Much Force is Needed to Keep a Drilled Cylinder at Rest?

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Homework Help Overview

The problem involves a uniform cylinder with a drilled hole, requiring the calculation of the horizontal force needed to keep it at rest. The subject area includes concepts of torque, center of mass, and static equilibrium.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need to determine the center of mass for the modified cylinder and explore different conceptualizations of the problem, including treating the drilled hole as a negative mass or considering symmetrical drilling.

Discussion Status

Some participants have offered alternative perspectives on how to approach the problem, suggesting different ways to visualize the mass distribution and torque calculations. There is an indication of productive dialogue, but no explicit consensus has been reached.

Contextual Notes

Participants note the importance of understanding the center of mass and the implications of torque in the context of static equilibrium. There may be constraints related to the assumptions about mass distribution and the geometry of the cylinder.

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Homework Statement


A uniform cylinder of radius R and mass 8kg has an off-axis hole drilled through it at 2R/5. Its new mass is 6.5kg. The hole and cylinder are parallel with their centers being the same height. What horizontal force, F, must be applied on the top to keep the cylinder at rest?


Homework Equations


I tried writing a torque equation for static equilibrium. However, I need to know the center of mass for the cylinder with the drilled hole, but I do not know it. Is there a better approach to this problem?


The Attempt at a Solution


Torque_net = (6.5)(distance of center of mass from cylinder's center) - (F)(R)=0
 
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Just think it is a solid cylinder plus a thin cylinder of negative mass.
 
Or you can imagine another hole, drilled symmetrically to the first, with the same mass taken out. The remaining cylinder will have 5 kg at x=0, one hole is actually filled with mass 1.5 kg at 2R/5. Then it shall be easy to calculate new xcom in terms of R.
Also g shall be involved in the torque calculation..
 
I got it, thanks so much!
 

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