I'm crap at physics - help!

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

Thanks guys!