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A small clamp of mass $m_{B}$ is attached at $B$ …

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Problem 113 Easy Difficulty

The center of gravity $G$ of a 1.5 -kg unbalanced tracking wheel is located at a distance $r=18 \mathrm{mm}$ from its geometric center $B$. The radius of the wheel is $R=60 \mathrm{mm}$ and its centroidal radius of gyration is $44 \mathrm{mm}$. At the instant shown, the center $B$ of the wheel has a velocity of $0.35 \mathrm{m} / \mathrm{s}$ and an acceleration of $1.2 \mathrm{m} / \mathrm{s}^{2},$ both directed to the left. Knowing that the wheel rolls without sliding and neglecting the mass of the driving yoke $A B,$ determine the horizontal force $\mathbf{P}$ applied to the yoke.


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Video Transcript

Hello friends as surrender figure there is unbalanced cracking bill off months. 1.5 kg on the radios up 60 millimeter having geometrical center and center off months. 18 millimeter away so that distance between the metrical center and center off mass or center of gravity will be 18. Limited radio central radius aberration is 40 form limited at that explain. So the center off will be having the velocity at given positions. Velocity off the is while 35 meter per second towards left on acceleration is also towards slept 1.2 m per second squared. We have to calculate the value of peace for you're rolling. Wish it first we will analyze the free body diagram off it. Mhm. Mhm. This is the B point geometrical center. This is the center off months. I hate having acceleration be towards left center Having the acceleration x Andi, if I value off by is R A B upon art and that off exit A B plus our Omega Chi Square acceleration, it can be defined at so we can write acceleration. It will be Evie plus our Omega Square towards left, unless are a B upon Capitola towards right. Sorry. Vertically apart. Now we have to throw the free body diagram. Maybe this is or gentle four street here the weight off the wheel will act. This will be normal reaction. And here it will be friction. Oh, it will be am a bar X. I am ever but calculating moment Top force about see point She is contact point with or gentle surface toe be into art Minus for me going toe are I am a bite into our I m a X and to our plus I into Alfa substituting the value Yeah, bait will be mgr I am a bye bye is our upon capital are into acceleration off the into r plus m h m Every palace our Amigas square are but as M K square. So this can be written as in the right hand side we can write mask Maybe you can take over Mars into every You will get r squared upon Capitola but smaller is square by art Bless Emma Here Omega will be the way they are unsolved ing it. This can be written. It's it's called two Can Britain m g are upon our m A. V is this a question to be one. Substituting the value mark off the really is given 1.5 g is 9.81 It's more art 18 limited, that is 180.0 bonnet capital are 60.6 mass 1.5 acceleration off the acceleration off. These given 1.2 small are quite zero bonnet in this radius of violation. 44 million to the point. Jiro Poor 40 square upon choir 2060 square Much 1.5 point 018 Why do you know 60 is square Where the city off These given Why do you fight on solving in while you were? Probably get 8.2 60 Newton towards left. That is the answer for this problem. Thanks for watching it.

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