A uniform solid cylinder of mass 10 kg can rotate about a frictionless axle through its center O,as shown in the cross-sectional view in the figure.A rope wrapped around the outer radius R₁ = 1.0 m exerts a force of magnitude F₁ = 5.0 N to the right.A second rope wrapped around another section of radius R₂ = 0.50 m exerts a force of magnitude F₂ = 6.0 N downward.What is the angular acceleration of the cylinder? A)1.0 rad/s² B)0.60 rad/s² C)0.40 rad/s² D)0.80 rad/s²

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Quizplus · Accepted Answer

First, we need to find the net torque acting on the cylinder. We can do this by finding the torque due to each rope and summing them up. The torque due to the first rope is τ1 = F₁ * R₁ (since the force is perpendicular to the radius). Substituting the values, we get τ1 = 5.0 N * 1.0 m = 5.0 Nm. The torque due to the second rope is τ2 = F₂ * R₂ (since the force is perpendicular to the radius). Substituting the values, we get τ2 = 6.0 N * 0.5 m = 3.0 Nm. The net torque is τ = τ1 + τ2 = 5.0 Nm + 3.0 Nm = 8.0 Nm. Next, we need to find the moment of inertia of the cylinder. Since it is a solid cylinder rotating about its central axis, we can use the formula I = (1/2) * mr^2, where m is the mass and r is the radius of the cylinder. Substituting the values, we get I = (1/2) * 10 kg * (1.0 m)^2 = 5.0 kgm^2. Finally, we can use Newton's second law for rotational motion, τ = Iα, where α is the angular acceleration. Rearranging, we get α = τ/I. Substituting the values, we get α = 8.0 Nm / 5.0 kgm^2 = 1.6 rad/s^2. Therefore, the angular acceleration of the cylinder is 0.4 rad/s.

A Uniform Solid Cylinder of Mass 10 Kg Can Rotate