A stick is resting on a concrete step with 2/7 of its length hanging over the ed

Question

A stick is resting on a concrete step with 2/7 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 41.3° 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 magnitude of the angular acceleration of the stick at the instant?

Explanation / Answer

here,

let the mass of the stick m.

length of the stick , l = 0.103 m

mass of ladybug , M = 2.92 * m

Using the standard formula, find the moment of inertia of the stick about its centre.

Using the parallel axis theorem, find the moment of inertia, I, of the stick about the pivot point.

I = m*l^2 /12 + m*(0.214 l)^2

I = m*0.103 *(2.92*(2/7)^2 + 2.92 * ( 5/7)^2 + 1/12 + 0.214^2)

I = 0.191 m kg.m^2

the torque (moment) produced by the lower bug. This is

2.92 * m *(2/7)*(0.103)cos(41.3)

Similarly the torque produced by the upper bug is
- 2.92 *m *(5/7)*(0.103)cos(41.3)

Find the net torque, t, by adding the 2 above values.

t = I * alpha

(2.92 * m *(2/7)*(0.103)cos(41.3) - 2.92 * m *(5/7)*(0.103)cos(41.3) ) *9.8 = 0.191 * m * alpha

alpha = 5.93 rad/s^2

the angular accelration is 5.93 rad/s^2

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