A stick is resting on a concrete step with 1/4 of its length hanging over the ed

Question

A stick is resting on a concrete step with 1/4 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 51.7

A stick is resting on a concrete with2/5 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 62.1 degree from the horizontal. If the mass of each bug is 3.09 times the mass of the stick and the stick is 12.7 cm long, what is the magnitude of the angular acceleration of the stick at the instant shown?

Explanation / Answer

moment of inertia of first bug = mr^2 = m * (2L/5)^2 = (4/25)3.09m L^2 = 0.4944 mL^2

moment of inertia of second bug = mr^2 = m * (3L/5)^2 = (9/25)3.09mL^2 = 1.1124 mL^2

moment of inertia of stick = (1/12) m L^2 + md^2 = (1/12) mL^2 + m (2.5L/5)^2 =

= (1/12 + 6.25 / 25) mL^2 = 0.33333 mL^2

total moment of inertia = (0.4944 + 1.1124 + 0.3333) mL^2 = 1.9401 mL^2

torque from first bug = - mgr cos62.1 = -3.26mg * (2L/5) cos62.1

torque from second bug = mgr cos62.1 = 3.26mg * (3L/5) cos62.1

torque from weight of stick = mgr cos46.5 = mg * (2.5L/5) cos62.1

total torque = (2.5 + 9.78 - 6.25) mg (L/5) cos62.1

= 1.206 mgL cos62.1

finally...

angular accelration = total torque / total moment on inertia

= 1.206 mgL cos62.1 / 1.9401 mL^2

= (1.206 * cos62.1 / 1.9401 g / L =

= 0.2908 * 9.8/ 0.127

= 22.445 rad/s^2

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