Question

Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 24 N, and F3 = 13 N. The radial distances are R2 = 0.22 m and R3 = 0.10 m. The forces F2 and F3 are perpendicular to the radial lines labeled R2 and R3. The moment of inertia of the cylinder is 1/2 MR 2/2. Find the magnitude of the angular acceleration of the cylinder about the axis of rotation.

Answer 1

Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

         ∑ τ = I α

         

we will assume that the counterclockwise turns are positive

         F₁  0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

        I = ½ M R₂²

         

        α = (F₂ R₂ - F₃ R₃)  $\frac{2}{M R_2^2}$

let's calculate

        α = (24  0.22 - 13  0.10) $\frac{2}{12 \ 0.22^2}$2/12 0.22²

        α = 13.7 rad / s²