By Dave Rappeport

With the hope that the 2012 NHL lockout may finally be coming to
an end and the coming approach of winter, I decided I should deepen my
understanding of the physics behind the game. More specifically, I want to
understand the physics behind the slapshot. Zdeno Chara of the Boston Bruins
currently holds the record for fastest slapshot at 108.8 mph, (48.6 m/s). How
fast must Chara’s stick rotate for the puck to have that speed? How much does
he bow his stick?

http://www.youtube.com/watch?v=bcieIx6_ArQ
(2 minute mark)

To simplify the problem:

-No friction

-No loss of velocity to
bowing the stick

The physics of the slapshot can be
summed up as an elastic collision whereby the kinetic energy in a hockey stick
and the momentum of the stick are transferred to the puck.

The official weight of an NHL puck is 6
oz, (0.17 kg). If we assume the puck starts at rest and the stick is only in
contact with the puck for 0.02s, we can use impulse to find the force exerted
on the puck.

SΔFt = Δp

SΔF(0.02s) = 0.17kg(48.6 m/s -0)

SFx = 413 N

Wnet = SFx

The ΔKE is made up of two parts:
rotational and translational kinetic energy. In order to calculate rotational
energy, let’s
model the stick as a long uniform rod rotating through its end. Therefore, the
radius is equal to the length. The average professional
defensemen’s stick is approximately 2.0 m in length.

How
fast is Chara’s stick rotating just before it hits the puck?

I= (1/3)ML^2

L = 2.0m = r

v = rω

vi = 0 = ωi

Wnet = 413 N

Wnet = ΔKE

413 N = (1/2)M*(vf^2-vi^2) + (1/2)*(1/3)ML^2*(ωf^2-ωi^2)

= (1/2)M*(rωf^2) + (1/2)*(1/3)ML^2*(ωf^2)

= .5(1.0)(4)*(ωf^2) +
(.17)(1.0)(4) *(ωf^2)

= 2.68*(ωf^2)

ωf = 12.4 rad/s

How much did Chara bow his stick?

First, we need the change in the spring
potential energy stored in the stick when the player bends (bows) it. Hockey
sticks have varying degrees of flex, the amount of force it takes to bend the
stick one inch. In the context of spring force, flex rating = k.

If we assume Chara’s stick has a flex
rating of 100 lbs, then k = -(440N/in), (11.2 N/m).

Now we need the change in gravitational
potential energy. In order to find it, we need the height of the end of the
stick during the player’s windup. 4Now let’s assume that the stick is not rotated a
full 180° (π) but about 75% of 180° (π), (3π/4). The height of the stick would
be given by the change in “l”, which = (3π/4)*(2.0m) = 4.7m.

Wnet = SFx

Wnet = 413 N

Wnet = ΔKE = -ΔPE

I= (1/3)ML^2

k = -(440N/in)

x = ?

h = 4.7m

413 N = Mgh + kx

= (1.0)(9.8)(4.7) + (-440)(x)

Good equation to solve this

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