A stick is resting on a concrete step with 1/3 of its total length 𝐿 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 mom...

Question

A stick is resting on a concrete step with 1/3 of its total length 𝐿 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 at 𝜃=67.3∘ with respect to the horizontal, as shown in the figure.

If the mass of each bug is 3.09 times the mass of the stick and the stick is 17.5 cm long, what is the magnitude 𝛼 of the angular acceleration of the stick at the instant shown? Use 𝑔=9.81 m/s2.

Answer 1

Let

L = 17.5 cm = 0.175 m be the length of the stick,

m (unknown) be the mass of the stick,

α (to be computed) be the angular acceleration.

When the stick is at an angle θ,

the horizontal distance of the first bug from the point of support is Lcos(θ)/3,

the horizontal distance of the second bug from the point of support is Lcos(θ)2/3,

The horizontal distance of the center of gravity from the point of support is Lcos(θ)/6.

Let's choose the direction of the rotation caused by the first bug to be negative,

and the direction of rotation caused by the second bug to be positive.

The first bug contributes a torque of -3.09mgLcos(θ)/3,

the second bug contributes a torque of 3.09mgLcos(θ)2/3, and

the weight of the stick contributes torque mgLcos(θ)/6

So the net torque is

τ = mgLcos(θ)(-3.09/3 + 3.09×2/3 + 1/6) = 7.18mgLcos(θ)/6

Assuming that the stick is uniform, its moment of inertia is

I = mL²/12

By Newton's second law

τ = Iα

So

α = τ/I = (7.18mgLcos(θ)/6) / (mL²/12) = 14.36gcos(θ)/L = 14.36×9.81×cos(67.3°)/0.175 = 310.6 s^-2