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A disk with a rotational inertia of 2.0 kg m2 and a radius of 1.6 m is free to rotate about a frictionless axis perpendicular to the disk's face and passing through its center. A force of 5.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk as a result of this applied force?
a) 4.0 rad/s2
b) 1.0 rad/s2
c) 2.0 rad/s2
d) 0.40 rad/s2
e) 0.80 rad/s2

Respuesta :

cjrodriguez89

Answer:

a) 4.0 rad/s2

Explanation:

  • For rigid bodies, Newton's 2nd law becomes :

       τ = I  * α (1)

        where τ is the net external torque applied, I is the rotational inertia

       of  the body with respect to the axis of rotation, and α is the angular

       acceleration caused by the torque.

  • At the same time, we can apply the definition of torque to the left side of (1), as follows:

       [tex]\tau = F*r*sin \theta (2)[/tex]

       where τ = external net torque applied by Fnet, r is the distance    

       between the axis of rotation and the line of Fnet, and θ is the

      angle between both vectors.

      In this particular case, as Fnet  is applied tangentially to the disk, Fnet

     and r are perpendicular each other.

  • Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:

       [tex]\alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2 (3)[/tex]        

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