A stick is resting on a concrete step with 2/11 of its length hanging over the e

Question

A stick is resting on a concrete step with 2/11 of its length hanging over the edge. A single ladybug lands on one 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 18.7 cm long, what is the magnitude of the angular acceleration of the stick at that instant?

Explanation / Answer

let m is the mass of rod

and L is the total length.

mass of Bug, M = 2.92*m

L = 18.7 = 0.187

moment of inertia of whole system about axis of rotation

I = I_stick + I_bugs

= m*L^2/12 + m*(9*L/11 - L/2)^2 + M*(2*L/11)^2 + M*(9*L/11)^2

= m*0.187^2/12 + m*(9*0.187/11 - 0.187/2)^2 + 2.92*m*(2*0.187/11)^2 + 2.92*m*(9*0.187/11)^2

= 0.078184*m

Torque acting on stick, Tnet = m*g*(9*0.187/11 - 0.187/2)*sin(90-41.3) + 2.92*m*g*(9*0.187/11)*sin(90-41.3) - 2.92*m*g*(2*0.187/11)*sin(90-41.3)

= m*9.8*(9*0.187/11 - 0.187/2)*sin(48.7) + 2.92*m*9.8*(9*0.187/11)*sin(48.7) - 2.92*m*9.8*(2*0.187/11)*sin(48.7)

= 2.996*m

now Apply, Tnet = I*alfa

==> alfa = Tnet/I

= 2.996*m/(0.078184*m)

= 38.3 rad/s^2 <<<<<<<<<<<<<------------------Answer

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