Dynamics Physics  Acceleration Analysis / Gear Ratios /by f2434427@rmqkr Tags: acceleration, analysis, dynamics, gears, physics 

#1
Jun1713, 06:49 AM

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First time posting.
1. The problem statement, all variables and given/known data Gears A and B, of Masses 4kg and 10kg, respectively, are rotating about their mass centers. The radius of gyration about the axis of rotation is 100mm for A and 300mm for B. A constant couple of C=0.75Nm acts on gear A. Neglecting friction, compute:  Angular Acceleration of each gear  Tangential contact force between the gears at C May want to look at attachment for image. 2. Relevant equations 3. The attempt at a solution Started with the smaller gear. Using T=I[itex]\alpha[/itex] , where I=mr^2 Rearrange the equation to get [itex]\alpha[/itex] = 0.75 / (4*0.1^2) = 18.75 rad/s^2 Not to sure how to draw a relationship with the 2nd larger gear. Maybe using w1/w2 = r2/r1 then [itex]\alpha[/itex] = w^2 * r Thanks, 



#2
Jun1713, 07:01 AM

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Welcome to PF;
The applied torque has to accelerate both gears doesn't it? If this were a linear case, you'd use a free body diagram to couple the masses wouldn't you? 



#3
Jun1713, 07:07 AM

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Ah yes, that would make sense. Forgot to account for the 2nd gears inertia
hmm. Dont quite understand what you mean by coupling the masses though 



#4
Jun1713, 07:41 AM

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Dynamics Physics  Acceleration Analysis / Gear Ratios /
When you push on one mass, it pushes on the other one .. the masses are said to be coupled ... the motion of one affects the motion of another.




#5
Jun1713, 07:52 AM

P: 3

Alrighty,
Well the force applied from gear A, will have an opposite reaction on Gear B. T=F*d F=0.75/0.15 = 5n then applying this force to Gear B. T=F*D = 5 * 0.45 = 2.25 Nm alpha = T/I = 2.25/(10*0.3^2) =2.5 rad/s^2 Looks about right to me..... hopefully. Logically, first gear would have to spin faster than the 2nd. 


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