Question

The trap-jaw ant can snap its mandibles shut in as little as 1.3 10-4 s. In order to shut, both mandibles rotate through a 90° angle. What is the average angular velocity of one of the mandibles of the trap-jaw ant when the mandibles snap shut in 2.20 10-4 s?

Answer 1

Answer

given,

Time,t = 1.3 x 10⁻⁴ s

angle of rotation = 90° = $\dfrac{\pi}{2}\ rad$

Average angular velocity of the mandible

$\omega =\dfrac{\Delta \theta}{\Delta t}$

$\omega =\dfrac{\dfrac{\pi}{2}}{1.3\times 10^{-4}}$

        ω = 1.21 x 10⁴ rad/s

average angular velocity is equal to 1.21 x 10⁴ rad/s

if the time given is = 2.20 x 10⁻⁴ s

$\omega =\dfrac{\Delta \theta}{\Delta t}$

$\omega =\dfrac{\dfrac{\pi}{2}}{2.20\times 10^{-4}}$

        ω = 7.14 x 10³ rad/s

average angular velocity is equal to 7.14 x 10³ rad/s

Answer 2

The average angular velocity of the mandibles is 7.14 x 10³ rad/s

Average angular velocity:

Given information:

Time taken Δt = 2.2 x 10⁻⁴ s

the angle of rotation Δθ = π/2  

The average angular velocity is defined as the rate of change of angular displacement.

Mathematically it is expressed as follows:

ω = Δθ/Δt

here Δθ is the angular displacement

and Δt is the time taken

so,

average angular velocity will be:

ω = (π/2) / (2.2 x 10⁻⁴)

ω = 7.14 x 10³ rad/s

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