A stick is resting on a concrete step with 1/6 of its length hanging over the ed

Question

A stick is resting on a concrete step with 1/6 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 degree from the horizontal. If the mass of each bug is 3.26 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 the instant shown? Number rad/s^2

Explanation / Answer

net torque = Mstick*g*(L/2 - L/6)*cos41.3 + Mbug*g*(L-L/6)*cos41.3 - Mbug*g*L/6*cos41.3

net torque = (Mstick*g*L/3 + Mbug*g*5L/6 - Mbug*g*L/6)8cos41.3

net torque = (Mstick*g*L/3 + Mbug*g*2L/3)*cos41.3

net torque = ((Mstick/3 + 3.26*Mstick*2/3)*L)*cos41.3

net torque = 1.89*Mstick*L

moment of inertia

I = (1/12)*Mstick*L^2 + (Mstick*(L/2 - L/6)^2) + Mbug*(L/6)^2 + Mbug*(L-L/6)^2

I = (1/12)*Mstick*L^2 + (Mstick*(L/3)^2) + Mbug*(L/6)^2 + Mbug*(5L/6)^2

I = (7/36)*Mstick*L^2 + Mbug*(26/36)*L^2

I = (Mstick*(7/36) + 3.26*Mstick*(26/36))*L^2

I = Mstick*((7/36) + (3.26*(26/36)))*L^2

I = 2.55*Mstick*L^2

net torque = I*alpha

1.89*Mstick*L = 2.55*Mstick*L^2*alpha

1.89 = 2.55*L*alpha

1.89 = 2.55*0.187*alpha

alpha = 3.96 rad/s^2 <<<<==========ANSWER

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