# Calculations

Dave Tutelman  --  Sept 1, 2014

Here is the detailed analysis, the mathematics, that is of less than general interest and would have interrupted the flow of the article.

## Impact velocity for different length levers

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:
F = τ / L          (where: τ (torque) = 120, and  L=20 or 40)

a = F/m         (derived from the basic F=ma and simple algebra. In this case, m=3)

D = π L
(the distance traveled is a half-circle of radius L)

D = ½ a t2     (the distance an accelerating body travels over time, which we solve to...)

t = √ 2D / a

V = at            (the velocity of an accelerating body over time)

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.

F = τ / L = ma

so

a = τ / mL

We also know

t = √ 2D / a
V = at

Which combine to give

V = √ 2 D a

Plugging in our values for D and a gives

V = √ 2 (K L) (τ / mL)

V = √ 2 K  τ / m)

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