EX25.1 - Revolving Door

EX25.1 - Revolving Door - linear momentum as well Force...

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Lesson 25 Example 25.1 Revolving Doors Revolving Door Another Revolving Door
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A person who walks through the revolving doors exerts a 90 N horizontal force on one of the doors. As he walks, he maintains the 15 degree angle relative to the line that is normal to the door panel. Each panel can be modeled as a 60 kg rectangular plate that is 1.2 m in length as viewed from above. Find the final angular velocity of the door if the person exerts the force for 3 seconds. The door is initially at rest and friction can be neglected. Deals with time, so using impulse momentum is a possibility. The door is a rigid body, so we will definitely need to use angular – lets start there, then we’ll see if we need to deal with
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Unformatted text preview: linear momentum as well. + Force Distance I o one bar I o = 2(1/12)(120)(2.4) 2 Sub in values. Could also break it into two bars, get the same I o . What type of problem? Find: Given: F = 90 N, m DP =60 kg, t=3 sec, θ = 15º ω DP EX25.1 – Revolving Doors Solution Reflection: Note this is about 17 rev/min. This is a reasonable angular velocity for a revolving door. We did not need to consider linear momentum in this case. Any forces at point O do not create a moment about O and therefore do not appear in the angular impulse-momentum equation. Note that for high speed air resistance will be quite large for a real door....
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This note was uploaded on 01/22/2011 for the course ME 212 taught by Professor Staff during the Spring '05 term at Cal Poly.

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EX25.1 - Revolving Door - linear momentum as well Force...

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