Question

A stick is resting on a concrete step with 2727 of its total length ?L 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 at ?=41.3∘θ=41.3∘ with respect to the horizontal, as shown in the figure.

If the mass of each bug is 3.093.09 times the mass of the stick and the stick is 16.3 cm16.3 cm long, what is the magnitude of the angular acceleration ?α of the stick at the instant shown?

Answer 1

From figure,

by net torque balance,

Torque is given by,

= FxR = F*R(perpendicular) = F(perpendicular)*R

So, net = Wb*(5/7)*L*cos(A) + Ws*(3/14)*L*cos(A) - Wb*(2/7)*L*cos(A)

here, A = angle = 41.3 deg

Wb = weight of bug

Ws = weight of stick

Given, Mb = 3.09*Ms

then, net = Ms*9.81*(3.09*5*0.163*cos(41.3 deg)/7 + 3*0.163*cos(41.3 deg)/14 - 3.09*2*0.163*cos(41.3 deg)/7)

net = 1.848276*Ms

now, net = I*

here, I = moment of inertia about hinge = Ib + Ib + Is(about hinge)

I = Mb*(2*L/7)^2 + Mb*(5*L/7)^2 + Ms*[(L^2)/12 + (3*L/14)^2]

I = 3.09*Ms*(2*0.163/7)^2 + 3.09*Ms*(5*0.163/7)^2 + Ms*[(0.163^2)/12 + (3*0.163/14)^2]

I = Ms*[3.09*(2*0.163/7)^2 + 3.09*(5*0.163/7)^2 + (0.163^2)/12 + (3*0.163/14)^2]

I = 0.052023*Ms

So, from above equation,

= net/I = 1.848276/0.052023

= 35.53 rad/sec.^2

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