I have a large gear having a moment of inertia about it's axis of
0.5Kgm2. If I wanted to accelerate it from static to a rotational
speed of 16 RPM in say, 2 seconds, what torque do I need to apply to a
driving pinion meshed with the gear? The gear ratio is 18:1. Assume
gear mesh efficiency at 100% for this argument.
Any offers?
Thanks

F=m*a (F=m*dv/dt)
translate to "angular world" for comparable formulation
keep careful track of your units
draw a free body diagram...including the gear geometry
find some other students to study / do homework with
get yourself a white board (dry erase)
I hope the homework's due on Monday not Friday but I'm betting on
Friday
If you want to learn mechanics you've got to do a fair number of
problems
get a study aid like Schaum's Outline.....still got mine somewhere
half.com

Example problem....A force of 5 N is applied to the rim of a bicycle
wheel 0.50 m in diameter. The mass of the wheel is 1.5 kg. What is the
angular acceleration of the wheel?
For straight line things,
Force = mass times acceleration,
and for rotating things,
torque equals moment of inertia (mass analog)
times
angular acceleration (acceleration analog)
Here, the moment of inertia is
I = mr2 = 1.5 kg x (0.25 m)2 = 0.0938 kg.m2
and the torque about the center of the wheel is
t = Fr = (5 N) x (0.25 m) = 1.25 N.m
so the angular acceleration
call it a(ang), of the wheel is
a(ang) = t/I = (1.25 N.m)/(0.0938 kg.m2) = 13.3 rad/s2
That example is not too hard to work through, I don't think.
So next, let's try YOUR problem.
Given, that.... moment of inertia
I = 0.5 kg.m2 and.... angular accel
a(ang) = 2 x pi x 16 / ( 2 x 60) = 0.838 rad/s2
...expressed as radians per second per second
So you can say pretty quickly, that the torque you
need on the gear's axis to spin it up is t = I times a(ang)
and that comes to
t = 0.5 kg.m2 times 0.838 rads/sec2 = 0.419 N.m
The meshing pinion needs to apply eighteen times
less torque if we neglect ITS moment of inertia,
and that would be 0.023 N.m
You can do this.....
Brian W

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