Question

The bunchberry flower has the fastest-moving parts ever observed in a plant. Initially, the stamens are held by the petals in a bent position, storing elastic energy like a coiled spring. When the petals release, the tips of the stamen act like medieval catapults, flipping through a 60^o­ angle in just 0.30 ms to launch pollen from anther sacs at their ends. The human eye just sees a burst of pollen; only high-speed photography reveals the details. As FIGURE CP12.85 shows, we can model the stamen tip as a 1.0mm long, 10 μg rigid rod with a 10 μg anther sac at the end. Although oversimplifying, we’ll assume a constant angular
acceleration.

a. How large is the “straightening torque”?
b. What is the speed of the anther sac as it releases its pollen?

Short answer

a) Straightening Torque is 29.52 x 10^-8 N.m

b) Speed of the anther sac as it releases its pollen is 7 m/sec

Step-by-step solution

Part(a) Step1 : Given information

length of rod = 1 mm = 0.001 m

mass = 10^-8 kg

time = 3 x 10^-4sec

angular displacement = 60^o = 1.0472 rads

Part(a) Step2 : Explanation

Angular acceleration is calculated by
$$\begin{gathered} \alpha =\frac{2\left(\text{displacement}\right(\theta \left)\right)}{t^2} \\ \alpha =\frac{2(1.0472rad)}{{\left(3\times {10}^{-4}s\right)}^2}=23.2\times {10}^{6}rad/s^2 \end{gathered}$$

We can calculate moment of inertia as

$I=\frac{1}{3}M{L}^2+M{L}^2$

$$\begin{gathered} I=\frac{1}{3}\left({10}^{-8}kg\right)(0.001m{)}^2+\left({10}^{-5}g\right)(0.001m{)}^2 \\ I=1.33\times {10}^{-14}kg.m^2 \end{gathered}$$

Calculate torque as $\tau =I\alpha$
Substitute values, we get

Torque $$\begin{gathered} =\left(1.33\times {10}^{-14}g.m^2\right)\times \left(23.2\times {10}^{6}rad/s^2\right) \\ =29.52\times {10}^{-8}\mathrm{Nm} \end{gathered}$$

Part(b) Step1 : Given information

length of rod = 1 mm = 0.001 m

mass = 10^-8 kg

time = 3 x 10^-4sec

angular displacement = 60^o = 1.0472 rads

Part(b) Step 2: Explanation

Angular velocity is given as

$$\begin{gathered} \omega =\alpha t \\ \omega =\left(23.2\times {10}^{6}rad/s^2\right)\left(3\times {10}^{-4}s\right) \\ \omega =6960\mathrm{rad}/\sec \end{gathered}$$

and velocity is v= ωL
On substitution we get
$\mathrm{v}=(6960rad/s)\times (0.001m)=6.960m/s=7\mathrm{m}/\sec$