Question

A stick is resting on a concrete step with 1 6 16 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 ?= 67.3 ∘ θ=67.3∘ with respect to the horizontal, as shown in the figure. If the mass of each bug is 3.09 3.09 times the mass of the stick and the stick is 11.5 cm 11.5 cm long, what is the magnitude ? α of the angular acceleration of the stick at the instant shown? Use ?=9.81 m/s 2 . g=9.81 m/s2.

Answer 1

From figure,

by net torque balance,

Torque is given by,

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

So, net = Wb*(5/6)*L*cos(A) + Ws*(1/3)*L*cos(A) - Wb*(1/6)*L*cos(A)

here, A = angle = 67.3 deg

Wb = weight of bug

Ws = weight of stick

Given, Mb = 3.09*Ms

L = length of stick = 11.5 cm = 0.115 m

then, net = Ms*9.81*(3.09*5*0.115*cos(67.3 deg)/6 + 1*0.115*cos(67.3 deg)/3 - 3.09*1*0.115*cos(67.3 deg)/6)

net = 1.04196*Ms

now, net = I*

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

I = Mb*(L/6)^2 + Mb*(5*L/6)^2 + Ms*[(L^2)/12 + (L/3)^2]

I = 3.09*Ms*(0.115/6)^2 + 3.09*Ms*(5*0.115/6)^2 + Ms*[(0.115^2)/12 + (0.115/3)^2]

I = Ms*[3.09*(0.115/6)^2 + 3.09*(5*0.115/6)^2 + (0.115^2)/12 + (0.115/3)^2]

I = 0.0320853*Ms

So,from above equation,

= net/I = 1.04196/0.0320853

= 32.47rad/sec.^2

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