A stick is resting on a concrete step with 27 of its length hanging over the edg

Question

A stick is resting on a concrete step with 27 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 com 46.500 momentarily to rest 46.5° from the horizontal. If the mass of each bug is 2.92 times the mass of the stick and the stick is 10.3 cm long, what is the nitude of the angular acceleration of the stick at the instant shown Number figure not to scale rad/s

Explanation / Answer

We will be writing the torque of the system about the point at which is in contact with table.

we know that,

Torque=(moment of inertia)(angualr accleration)

Let the mass of the stick be 'm' so mass of the bug will be 2.92m

So,

Moment of inertia of system,

=(2.92m*(2*.103/7)2)+(2.92m*(5*.103/7)2)+(m*(3*.103/14)2)+(m*(.103)2/12)

=(2.528*10-3+0.0158+5.967*10-3+8.84*10-4)m

=0.025179m

And torque is,

T=[2.92mg*(5L/7)cos46.5]+[mg*(3/14)L]-[2.92mg*(2L/7)cos46.5]

T=1.45m+0.2165m-0.58027m

=1.0862m

So,

1.0862m=0.025179m*angular accelration

angular accelration=43.14 rad/s2

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