A ruler is resting on the edge of a ledge, with 1/6 of its length hanging over t

Question

A ruler is resting on the edge of a ledge, with 1/6 of its length hanging over the edge. A single beetle lands on the end of the ruler hanging over the edge, and the ruler begins to tip. A moment later, a second, identical beetle lands on the other end of the ruler, which results in the ruler coming momentarily to rest 41.3degree from the horizontal. If the mass of each bug is 2.75 times the mass of the ruler and the ruler is 15.1 cm long, what is the magnitude of the angular acceleration of the ruler at the instant shown? Number = rad/s^2

Explanation / Answer

the moment the stick comes to rest at =41.3° from horizontal.

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

_lowerbug = (1/6)L(2.75mg)cos (negative because I am choosing counter-clockwise as the negative angular direction).

The upper ladybug exerts this much torque:

_upperbug = +(5/6)L(2.75mg)cos

The weight of the stick can be assumed to act through its center, which is 1/3 of the way from the fulcrum. So the stick exerts this much torque:

_stick = +(1/10)L(mg)cos

The net torque is thus:

_net = _lowerbug + _upperbug + _stick
= (1/6)L(2.75mg)cos + +(5/6)L(2.75mg)cos + (1/3)L(mg)cos
= (2.75(5/6(1/6))+1/10)(mgL)cos

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)((1/6))L)² = (2.75m)(1/36)L²
I_upperbug = (2.75m)((5/6))L)² = (2.75m)(25/36)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=(1/3)L. So:

I_stick = (1/12)mL² + m((1/3)L)²
= (1/12)mL² + (1/9)mL²
= (7/36)mL²

So the total moment of inertia is:

I_total = I_lowerbug + I_upperbug + I_stick
= (2.75m)(1/36)L² + (2.75m)(25/36)L² + (7/36)mL²
= (2.75(1/36+25/36)+7/75)mL²

So that means the angular acceleration is:

= _net/I_total
= (2.75(5/6(1/6))+1/10)(mgL)cos /(2.75(1/36+25/36)+7/75)mL²

The "m" cancels out.
= 1.933*g*cos41.3°/(2.079*L)
= 1.933*9.8*cos41.3°/(2.079*15.1)
= 0.4533

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