Genius Insect!

Let us take an overview of the problem

1. This problem is analogous to 'Variable Mass' problems in Translational motion.
2. Mass change in translational motion & inertia change here is similar.
3. Inertia changes as the insect moves from the point where it falls and travels till the end.

• First we find the angular velocity $omega$ with which this rod is going to rotate
• By the time the bug reaches the rod it will have $\sqrt{2gh}$ velocity.
• Applying conservation of angular momentum about hinge point, we get

$\frac{mvl}{4} = [m \times (l/4)^{2} + \frac{m \times l^{2}}{12}] \times \omega$

1. Now as discussed we get a torque due to changing moment of inertia
2. This torgue will be equal to torque due to the mass of the insect at that instant

$T = \frac{dI}{dt}w + \frac{dw}{dt}I$

$T = \frac{dI}{dt}w$

$T = 2mr\frac{dr}{dt}\omega$

Also at any given $\theta$ , torque due to insect's weight is $mg \cos \theta \times r$

• Here r is the distance from hinge to insect at any time t.
• Also $\theta = \omega \times t$
• The net torgue on the rod is zero hence the differential equation we get is

$mg \cos \theta \times r = 2mr\frac{dr}{dt}\omega$

Integrating this and substituting limits as

• Time: $[0] to [\frac{\pi}{2\omega}]$
• Distance: $[l/4] to [l/2]$
• Also substituting the value from COAM equation, we get

$h = \boxed{1.701}$

Apply conservation of angular momentum about hinge point & find w This 'w' will remain constant. Now this problem is analogous to 'Variable Mass' problems in Translational motion. Mass change in translational motion & inertia change here is similar. Torque=(dI/dt)w+I(dw/dt) =(dI/dt)w At any time 't' let distance of bug from hinge be' r' Hence, Torque=[2mr(dr/dt)]w Now due insect's mass Another torque acts (in opposite direction) As net torque=0 we have: 2mr(dr/dt)=mg[cos(wt)] where 'wt' is the angle subtended at that time t w.r.t. vertical integrate using the limits of time as: 0 to pi/2w and r as: l/4 to l/2 then use the relation we obtained from conservation of ang momentum you will get it as w=12v/7l v is insect's velocity when it strikes the rod Hence sqrt(2gh)=v substitute

Apply conservation of angular momentum about hinge point & find w This 'w' will remain constant. Now this problem is analogous to 'Variable Mass' problems in Translational motion. Mass change in translational motion & inertia change here is similar. Torque=(dI/dt)w+I(dw/dt) =(dI/dt)w At any time 't' let distance of bug from hinge be' r' Hence, Torque=[2mr(dr/dt)]w Now due insect's mass Another torque acts (in opposite direction) As net torque=0 we have: 2mr(dr/dt)=mg[cos(wt)] where 'wt' is the angle subtended at that time t w.r.t. vertical integrate using the limits of time as: 0 to pi/2w and r as: l/4 to l/2 then use the relation we obtained from conservation of ang momentum you will get it as w=12v/7l v is insect's velocity when it strikes the rod Hence sqrt(2gh)=v substitute