In the discussion of leverage, we compared two cases. Both had an applied torque of 120 at one end of the lever, and a mass of 3 at the other end. In one case the length of the lever was 20, the other case was 40. (Don't worry about units; we're just keeping numbers simple and assuming some consistent set of units.)

Let's assume the torque rotates the lever from a dead stop through 180º, a half circle. We want to know the finishing velocity of the mass.

The relevant formulas here are:
Applying these formulas in order for the two cases:

L = 20,  τ = 120,  m = 3	L = 40,  τ = 120,  m = 3
F = τ / L =  120 / 20  =  6 a = F/m  = 6 / 3  =  2 D = π L  =  3.14 * 20  =  62.8 t = √ 2D / a   =  sqrt( 2 * 62.8 / 2 )  =  7.92 V = at  =  2 * 7.92  =  15.8	F = τ / L =  120 / 40  =  3 a = F/m  = 3 / 3  =  1 D = π L  =  3.14 * 40  =  125.6 t = √ 2D / a   =  sqrt( 2 * 125.6 / 1 )  =  15.8 V = at  =  1 * 15.8  =  15.8

So either way, the velocity "at impact" is 15.8. Leverage does not help, nor does it hurt.

This result does not depend on the values we chose for lever length, mass, torque, nor the angle through which we rotate. We are rotating through a distance of πL, but we could set D to some arbitrary multiple of L (say, D = KL), and the impact velocity will still be completely independent of L. Let's go ahead and do the proof. We will start with the formulas above, and solve for V.


so


We also know


Which combine to give


Plugging in our values for D and a gives


Look, ma! No L in the formula for velocity.