Question

A stick is resting on a concrete step with 1/7 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 36.1° from the horizontal. If the mass of each bug is 2.75 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 shown?(answer in rad/s^2)

Answer 1

at the moment the stick comes to rest at theta = 36.1^o from horizontal.

Angular acceleration = (net torque) / (moment of inertia)

Here 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.

Then, 2.75*m is each bug's mass.

And let's say the "lower" ladybug is on the left.

Then the lower ladybug exerts a torque of : t_lower = −(1/7)*L*2.75*m*g*Cos(theta)

(Here negative is taken because I am (arbitrarily) choosing counter-clockwise as the negative angular direction).

t_upper = + (6/7)*L*2.75*m*g*Cos(theta)

The weight of the stick can be assumed to act through its center, which is 0.357 L of the way from the fulcrum.

So the stick exerts a torque: t_stick = +0.357*L*m*g*Cos(theta)

The net torque is thus:

t_net = t_lower + t_upper+ t_stick

t_net = −(1/7)L(2.75mg)cos(theta) + (6/7)L(2.75mg)cos(theta) + (0.357)L(mg)cos(theta)

t_net = (2.32)(mgL)cos(theta)

Now for the moments of inertia. The bugs can be considered point masses of "2.75*m" each.

So for each of them you can use the simple formula:

I = mass*R²;

I_lower = (2.75*m)((1/7)L)² = 0.056 mL²

I_upper = (2.75*m)((6/7)L)² = 2.02 mL²

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 = (1/12)mL²

And in this problem we know that it's offset by R = (0.357)L

So;

I_stick = (1/12)mL² + m((0.357)L)² = 0.21 mL²

So the total moment of inertia is:

I_total = I_lower + I_upper + I_stick

I_total  = ( 0.056 + 2.02 + 0.21) * mL²

I_total  = 2.286 mL²

So that means the angular acceleration is:

alpha = {2.32*m*g*L*Cos36.1^o} / (2.286*m*L²)

alpha = 18.37/0.4275

The magnitude of the angular acceleration of the stick at hte instant shown is, alpha = 42.97 rad/s²