A disk with a rotational inertia of 1.4 kg×m2 and a radius of 0.8 m rotates on a frictionless fixed axis perpendicular to the disk faces and through its center. A force of 5.2 N is applied tangentially to the rim. The angular acceleration of the disk is:
Torque = I alpha
5.2 * .8 = 1.4 alpha
alpha = 2.97 radians/second^2
To find the angular acceleration of the disk, we can use the equation:
τ = I * α
where τ is the torque applied, I is the rotational inertia of the disk, and α is the angular acceleration.
Given that the torque applied is τ = 5.2 N, and the rotational inertia is I = 1.4 kg·m^2, we can rearrange the equation to solve for α:
α = τ / I
Substituting the values into the equation, we have:
α = 5.2 N / 1.4 kg·m^2
Simplifying the equation, we get:
α = 3.71 rad/s^2
Therefore, the angular acceleration of the disk is 3.71 rad/s^2.