What force must be exerted by the wind to support a 2.50 kg chicken in the posit

Question

What force must be exerted by the wind to support a 2.50 kg chicken in the position shown below? Use the pivot point as the chicken's foot, the height of the wind as 20 cm, and the center of mass of the chicken as 9.0 cm to the left of the pivot. The moment of inertia of an ice skater is 0.400 kg middot m^2 when he is spinning at 6.00 rev/s. (a) He reduces his angular velocity by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (b) Suppose instead he keeps his arms in and allows friction on the ice to slow him to 3.00 rev/s What average torque was exerted if this takes 15.0 s?

Explanation / Answer

6). Applying equilibrium condition about the chicken's foot,

Fwind* distance= Moment about the point of contact with ground

Fwind* distance= Weight of chicken* distance from chicken's foot

  using all given values in SI units in above,

Fwind* (0.2)= (2.50)(9.8)(0.09)

Fwind= 11.025 N (ANS)

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