Warning - formulas ahead ! Skip if you are allergic...

For more technically inclined I can put numbers on those figures to better illustrate my point. Lets say elliptical stylus has contact surface of 25 μm² and it creates frictional force of 0.001N. Now average platter, with diameter of 310mm, weighting 2.2 kg and freewheeling at 33rpm, has inertia of 0.026kg·m² . To stop it in 1 second, you’d have to apply frictional torque T of :

$T=I*a=I*\frac{[-\omega]}{t}$ $\omega =\frac{(2*Pi)}T{}=2*Pi*f$

$\omega =2*3.14*(33/60)=3.45 \, rad\, s$

$T=0.026*[-3.45]/1=0.091 Nm$

$\small I \,- moment \, of \, inertia \, (kg\, m^2)$

$\small \omega \, -angular\, velocity\, (rad\, s)$

$\small a\, -angular\, acceleration \, (rads/s^2)$

$\small \small f - frequency \, of \, platter \, rotation$

If we assume that we are braking in a middle of a record, 9cm from center, this would translate to force F of :

$F=T/r=0.091/0.09\approx 1N$

$r-radius\, of\, applied\, force (m)$

And even if we are very generous and assume that stylus frictional force varies in a range of 10%, which translates to 0.0001N, that’s still a forces ratio of 1 to 10000 ! You wonder how this 0.0001N variation influences speed ? Again, lets be generous and apply this force for 1 second and work our way backwards:

$\omega =\frac{T*t}{I}=\frac{0.0001*1}{0.091}=0.01 \, rad\, s$

$rpm=\frac{\omega }{2*Pi}*60=\frac{0.01}{2*3.14}*60=0.095\,$

$rpm_{change}=\frac{33}{0.095}=0.03\, \%$

But you know what’s the funny part ? To fully counter this speed change of 0.03%, table motor drive must provide same minuscule 0.0001N of torque. Well, you get the idea now…