L=2/9 & 56.9 degrees from horizontal & mass of each bug is 2.75times the mass of

Question

L=2/9 & 56.9 degrees from horizontal & mass of each bug is 2.75times the mass of the stick when the stick is 17.5cm long....what is the magnitude of the angular acceleration of the stick at the instant shown? Answer in rad/s^2
O www.saplinglearning.com/ibiscms/mod/lbis/view.php?id-6173442 apling Learning INCLUSIVE ACCESS) and Angular Jump to.. Gabriel Zepeda 29/2018 11:59 PM 81/100 ?Print lakulator end Periode Table ion 3 of 12 Sapling Learning is resting on a concrete step with 2/9 of its hanging over the edge. A single ladybug 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 comin momentarily to rest 56 9 from the mass of each bug is 2.75 times the mass of the stick and the stick is 17.5 cm long, what is the magnitude of the at the instant shown 56900 f the acceleration of the stick figure not to scale

Explanation / Answer

At the moment the stick comes to rest at (th)= 56.9° from horizontal.

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

alpha = T/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:

T_lowerbug = ?(2/9)L(2.75mg)cos(th) (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction).

The upper ladybug exerts this much torque:

T_upperbug = +(7/9)L(2.75mg)cos(th)

The weight of the stick can be assumed to act through its center, which is 2.5/9 = 5/18 of the way from the fulcrum. So the stick exerts this much torque:

T_stick = +(5/18)L(mg)cos(th)

The net torque is thus:

T_net = T_lowerbug + T_upperbug + T_stick

= ?(2/9)L(2.75mg)cos(th) + (7/9)L(2.75mg)cos(th) + (5/18)L(mg)cos(th)

= (2.75(-2/9+7/9)+5/18)(mgL)cos(th)

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)((2/9)L)² = (2.75m)(4/81)L²

I_upperbug = (2.75m)((7/9)L)² = (2.75m)(49/81)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. And in this problem we know that it's offset by R=(5/18)L. So:

I_stick = (1/12)mL² + m((5/18)L)²

= (1/12)mL² + (25/324)mL²

= (13/81)mL²

So the total moment of inertia is:

I_total = I_lowerbug + I_upperbug + I_stick

= (2.75m)(4/81)L² + (2.75m)(49/81)L² + (13/81)mL²

= (2.75(4/81+49/81)+13/81)mL²

So that means the angular acceleration is:

alpha = T_net/I_total

= ((2.75(-2/9+7/9)+5/18)(mgL)cos(th) )/( (2.75(4/81+49/81)+13/81)mL² )

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

alpha = 28.17 rad/s^2

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