determine: (a) the angular velocity of the gate immediately after the collision, and (b) the maximum angular displacement of the gate, θ (in radians)

Question

A car trying to break into a secure industrial area collides with a spring-

loaded “swing away” security gate. The gate can be modeled as a bar with a torsional

spring mounted at the pivot point.

The mass moment of inertia of the bar about the axis through point O is I_O = 900 kg-m2,

and the torsional stiffness of the spring is

k = 4500 N-m. The car has a mass m = 1965 kg

and hits the gate at point P,2.4 m from pivot point O.

If the velocity of the car just prior to collision is v = 30 m/s and the coefficient of restitution

of the collision, e = 0.71, determine: (a) the angular velocity of the gate immediately after

the collision, and (b) the maximum angular displacement of the gate, θ (in radians).

Answer 1

Let

L = 2.4 m be the distance from pivot O to the impact point,

v = 30 m/s be the velocity of the car before the collision,

u (unknown) be the velocity of the car after the collision,

w (unknown) be the velocity of the point P after the collision,

ω (to be calculated) be the angular velocity of the gate after the collision.

The initial direction of the car towards the gate is considered positive,

while the opposite direction is negative.

(a)

For question (a) we use conservation of angular momentum and

the definition of coefficient of restitution to find velocities u and w.

mvL is the angular momentum (with respect to the point O) of the car before the collision,

muL is the angular momentum (with respect to the point O) of the car after the collision,

Iω is the angular momentum (with respect to the point O) of the gate after the collision.

Using ω = w/L, and using conservation of angular momentum,

mvL = muL + Iw/L (1)

From the definition of coefficient of restitution

e = (w - u)/(v - 0) (2)

From (2)

u = w - ev (3)

Substitute (3) into (1)

mvL = m(w - ev)L + Iw/L

From that

w = mLv(1 + e)/(mL + I/L)

By definition of angular velocity

ω = w/L = mv(1 + e)/(mL + I/L)

Substituting actual numbers

ω = 1965×30×(1 + 0.71)/(1965×2.4 + 900/2.4) = 19.8/s

(b)

Question (b) is answered from conservation of energy,

assuming no friction in the movement of the gate.

Right after the collision the gate has kinetic energy Iω²/2.

At maximum displacement the initial kinetic energy got converted to the spring energy kθ²/2.

Setting them equal

kθ²/2 = Iω²/2

θ = ω√(I/k)

Substituting actual numbers

θ = 19.8√(900/4500) = 8.85

That means that the gate makes more than a full circle before stopping.

That should be hardly surprising.

The problem describes an SUV exceeding highway speeds hitting a gate weighing less than a golf cart.

It hardly slows down the SUV:

from v = 30 m/s to u = 26 m/s