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A uniform rod is set up so that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod are 0.899 m0.899 m and 1.13 kg1.13 kg , respectively. A force of constant magnitude FF acts on the rod at the end opposite the rotation axis. The direction of the force is perpendicular to both the rod's length and the rotation axis. Calculate the value of FF that will accelerate the rod from rest to an angular speed of 6.61 rad/s6.61 rad/s in 8.97 s8.97 s .

Respuesta :

Answer:

Explanation:

angular acceleration α = 6.61 / 8.97 = .737 rad / s²

moment of inertia = 1/3  m L²

= 1/3  x 1.13 x .899²

= .3045  kg m²

torque applied = moment of inertia x angular acceleration

= .737 x .3045 = .2244

If be the force

Torque

= F x .899 = .2244

F = .25 N ,

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