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 halfcircle of radius L)
D = ½ a t^{2}
(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.