A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius R1 = 1.0m, exerts a force F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force F2 = 6.0 N downward. What is the angular acceleration of the disk? ...?

Torque acting dowward = 6 x 0.5 = 3 Nm

Torque acting to the right = 5 x 1 = 5 Nm

5 - 3 = 2 Nm

inertia = 1/2 mr^2

0.5 x 10 x 1^2 = 5 kg-m^2
2/5 = alpha = 0.4 rad /s^2

Hope this helps

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Lucianorayos1 Lucianorayos1 · Answer 2 · 2019-09-12T18:33:30-04:00

Answer:

∝=0,4\(s^{-2}\) (angular acceleration)

Explanation:

We give you the equation of the sum of all torques:

∑τ=I∝

Then we replace the torques :

F1R1-F2R2=I∝ The inertia of a cylinder mass that pivot arround the axis in the center of it is: \(\frac{MR1^{2}}{2}\), so then we can replace. don't confuse the R1 of the intertia of the cylinder, with the R2 of the F2. And also you must consider the direction of forces, and your reference system, in this case i choose clockwise, that's why my F1 is possitive, and my F2 is negative.
\(\frac{F1*R1-F2*R2}{I} =\alpha\)

\(\frac{5N*1M-6N*0,5M}{\frac{10KG*1M^{2} }{2} }= \alpha\)

\(\frac{2*(5N.M-3N.M)}{10KG*1M^{2} }= \alpha\)
\(\frac{(2N.M)}{5KG*1M^{2} }= \alpha\)

\(\alpha =0,4s^{2}\)