Key Leads and Inertia Experiment

[Phillip Ford]

>Where r =10 cm; rb = 5 cm; rk = 2.5 cm;  m = 30 grams and M = 140 grams. 
>(just accept these as working values)
>
>whats going to be our breakpoint acceleration, and what will be the force 

>needed to get there ?
>
>Before trying to work this out... I took M of 140 grams because thats a 
>fairly typical weight that could be expected to be resting on the capstan 

>of a key in the middle / low part of a piano. I'm unsure as to whether 
>this is right  to insert this here...... so correct if wrong.
>
>In anycase.... g is the acceleration due to gravity right ?... ie. circa 
>9.8 ms^2 ?
>
>That gives if so (r/rb)g =  19.6ms^2 which should then be the breakpoint 
>acceleration for this example.
>
>Fbr then is (140 * 2.5 + Ik/5 )( 9.8/10), and if Ik can be understood to 
>be Mrk^2 then thats 140 * 6.25 = 875 which ends up yielding a force of 
>514.5 grams.
>
>Which if even remotely correct would mean we are confined to pretty light 

>play levels for this soft zone.
>
>Ok .... this should suffice to show exactly where I am confused about
what 
>to plug in where in the formulas at the appendix. So... please straighten 

>me out on this one :)
>
>RicB

Ric,

This all looks OK except that you need to be consistent in your units. For your distance measurements you're using cm, but for g you are using m/s^2.  You need to be using 980 cm/s^2 or change your distance measurements to meters.

Phil F