Map Sapling Learning macmillan leanning A stick is resting on a concrete step wi
ID: 1795170 • Letter: M
Question
Map Sapling Learning macmillan leanning A stick is resting on a concrete step with 1/5 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 a second, identical ladybug lands on the other end of the stick, which results in the stick comin momentarily to rest 67.3° from the horizontal. If the mass of each bugis 2.75 times the mass of the stick and the stick is 12.7 cm long, what is the later 67.300 al the instant shoanqular acceleration of the stick Number figure not to scale rad/sExplanation / Answer
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 = (1/5)L(2.75mg)cos (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction).
The upper ladybug exerts this much torque:
_upperbug = +(4/5)L(2.75mg)cos
The weight of the stick can be assumed to act through its center, which is 3/10 of the way from the fulcrum. So the stick exerts this much torque:
_stick = +(3/10)L(mg)cos
The net torque is thus:
_net = _lowerbug + _upperbug + _stick
= (1/5)L(2.75mg)cos + (4/5)L(2.75mg)cos + (3/10)L(mg)cos
= (2.75(4/51/5)+3/10)(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)((1/5)L)² = (2.75m)(1/25)L²
I_upperbug = (2.75m)((4/5)L)² = (2.75m)(16/25)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=(3/10)L. So:
I_stick = (1/12)mL² + m((3/10)L)²
= (1/12)mL² + (9/100)mL²
= (13/75)mL²
So the total moment of inertia is:
I_total = I_lowerbug + I_upperbug + I_stick
= (2.75m)(1/25)L² + (2.75m)(16/25)L² + (13/75)mL²
= (2.75(1/25+16/25)+13/75)mL²
So that means the angular acceleration is:
= _net/I_total
= (2.75(4/51/5)+3/10)(mgL)cos / (2.75(1/25+16/25)+13/75)mL²
= (2.75(4/51/5)+3/10)(g)cos / (2.75(1/25+16/25)+13/75)L
= (2.75(4/51/5)+3/10)(9.8)cos67.3 / (2.75(1/25+16/25)+13/75)0.127
= 28.42 rad/s2