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

Question

Quizplus · Accepted Answer

We can use the formula for torque, τ = Fr, where F is the force applied and r is the distance from the pivot point. We need to find the total torque acting on the cylinder.

For the force of 5.0 N, the distance from the pivot point is 1.0 m, so the torque is 5.0 N * 1.0 m = 5.0 Nm.

For the force of 6.0 N, the distance from the pivot point is 0.50 m, so the torque is 6.0 N * 0.50 m = 3.0 Nm (note that we need to take the direction of the force into account, so when the force is downward, the torque is upward).

The total torque is then 5.0 Nm - 3.0 Nm = 2.0 Nm.

Next, we can use the formula for rotational motion, τ = Iα, where I is the moment of inertia and α is the angular acceleration. We need to find the moment of inertia of a solid cylinder about its center, which is given by I = (1/2)MR^2. Here, M is the mass of the cylinder and R is the radius.

Plugging in the given values, we get I = (1/2)(10 kg)(0.5 m)^2 = 1.25 kg m^2.

Finally, we can solve for α:

τ = Iα
2 Nm = 1.25 kg m^2 * α
α = 2 Nm / (1.25 kg m^2) = 1.6 rad/s^2

So the angular acceleration of the cylinder is 1.6 rad/s^2, which is closest to choice C, 0.40 rad/s^2.

A Uniform Solid Cylinder of Mass 10 Kg Can Rotate