Question

A trap-jaw ant has mandibles that can snap shut with some force, as you might expect from its name. The formidable snap is good for more than capturing prey. When an ant snaps its jaws against the ground, the resulting force can launch the ant into the air. Here are typical data: An ant rotates its mandible, of length 1.30 mm and mass 130 μg(which we can model as a uniform rod rotated about its end), at a high angular speed. As the tip strikes the ground, it undergoes an angular acceleration of 3.5×10^8 rad/s^2.

If we assume that the tip of the mandible hits perpendicular to the ground, what is the force on the tip?

How does this compare to the weight of a 12 mg ant?

Answer 1

We know that

$\tau= I \alpha$

where,

$\tau= torque$

$I=$ Moment of Inertia

$\alpha=$ angular acceleration

Now, if "L" is the length of the rod considered then

$L \times F= \frac{ML^2}{3}\times \alpha$

Since moment of inertia is considered about the end of the rod,

$FLsin90^\circ= \frac{ML^2}{3} \alpha$

$F= \frac{ML}{3} \alpha$

Using all given values in above,

$F= \frac{130 \times 10^{-9} kg \times 1.30 \times 10^{-3}}{3} \times 3.5 \times 10^8= 0.0197N$(ANS)

Answer 2