Question

A stick is resting on a concrete step with 2/5 of its length hanging over the edge. A single ladybug lands on the end of the stick hanging over the edge, and the stick begins to tip. A moment later, a second, identical ladybug lands on the other end of the stick, which results in the stick coming momentarily to rest 41.3° from the horizontal. If the mass of each bug is 3.43 times the mass of the stick and the stick is 18.7 cm long, what is the magnitude of the angular acceleration of the stick at the instant?

Answer 1

The moment the stick comes to rest at θ=62.1° from horizontal. 

Angular acceleration = (net torque) / (moment of inertia) 
α = τ/I 

We have to add up the torques due to the bugs and the stick; and add up the moments of inertia due to all three also. 

Let L be the stick's length and let m be the stick's mass (so "2.75m" is each bug's mass). And let's say the "lower" ladybug is on the left. Then the lower ladybug exerts this much torque: 

τ_lowerbug = −(2/5)L(2.75mg)cosθ (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction). 

The upper ladybug exerts this much torque: 

τ_upperbug = +(3/5)L(2.75mg)cosθ 

The weight of the stick can be assumed to act through its center, which is 1/10 of the way from the fulcrum. So the stick exerts this much torque: 

τ_stick = +(1/10)L(mg)cosθ 

The net torque is thus: 

τ_net = τ_lowerbug + τ_upperbug + τ_stick 
= −(2/5)L(2.75mg)cosθ + (3/5)L(2.75mg)cosθ + (1/10)L(mg)cosθ 
= (2.75(3/5−2/5)+1/10)(mgL)cosθ 

Now for the moments of inertia. The bugs can be considered point masses of "2.75m" each. So for each of them you can use the simple formula: I=mass×R²: 

I_lowerbug = (2.75m)((2/5)L)² = (2.75m)(4/25)L² 
I_upperbug = (2.75m)((3/5)L)² = (2.75m)(9/25)L² 

For the stick, we can use the parallel axis theorem. This says, when rotating something about an axis offset a distance "R" from its center of mass, the moment of inertia is: 

I = I_cm + mR² 

We know that for a stick about its center of mass, I_cm is (1/12)mL² (see many sources). And in this problem we know that it's offset by R=(1/10)L. So: 

I_stick = (1/12)mL² + m((1/10)L)² 
= (1/12)mL² + (1/100)mL² 
= (7/75)mL² 

So the total moment of inertia is: 

I_total = I_lowerbug + I_upperbug + I_stick 
= (2.75m)(4/25)L² + (2.75m)(9/25)L² + (7/75)mL² 
= (2.75(4/25+9/25)+7/75)mL² 

So that means the angular acceleration is: 

α = τ_net/I_total 
= ((2.75(3/5−2/5)+1/10)(mgL)cosθ)/((2.75(4... 

The "m" cancels out. You're given "L" and "θ" and you know "g", so do the math (and don't forget to use consistent units).