A hollow cylinder of outer radius R and mass M with moment of inertia about the center of mass Icm=MR2 starts from rest and moves down an incline tilted at an angle θ from the horizontal. The center of mass of the cylinder has dropped a vertical distance h when it reaches the bottom of the incline. Let g denote the acceleration due to gravity. The coefficient of static friction between the cylinder and the surface is μs. The cylinder rolls without slipping down the incline. The goal of this problem is to find an expression for the smallest possible value of μs such that the cylinder rolls without slipping down the incline plane and the velocity of the center of mass of the cylinder when it reaches the bottom of the incline.

(a) What is the magnitude of the acceleration a of the center of mass of the cylinder on the incline? Express your answer in terms of θ and g as needed (enter theta for θ and g for g).

a=

unanswered
(b) What is the minimum value for the coefficient of static friction μs such that the cylinder rolls without slipping down the incline plane? Express your answer in terms of θ (enter theta for θ).

μs min=

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(c) What is the magnitude of the velocity of the center of mass of the cylinder when it reaches the bottom of the incline? Express your answer in terms of g and h as needed (enter g for g and h for h).

vf=

http://web.mit.edu/8.01t/www/materials/ProblemSets/Raw/old_files_fall07/ps10sol.pdf

To solve this problem, we need to break it down into smaller steps. Let's begin by considering the forces acting on the cylinder as it moves down the incline.

First, there is the gravitational force pulling the cylinder downward. This force can be decomposed into two components: one parallel to the incline and one perpendicular to the incline.

The component of the gravitational force parallel to the incline, F_par, can be calculated as F_par = m * g * sin(θ), where m is the mass of the cylinder and g is the acceleration due to gravity.

Next, we have the static friction force, F_friction, opposing the motion of the cylinder. The maximum possible value of static friction, F_friction_max, is given by F_friction_max = μs * F_normal, where μs is the coefficient of static friction and F_normal is the normal force exerted by the incline on the cylinder.

In this case, the normal force, F_normal, is equal to the weight of the cylinder, which is given by F_normal = m * g * cos(θ).

Since the cylinder is rolling without slipping, the friction force can be related to the linear acceleration of the center of mass, a_cm, and the radius of the cylinder, R, by F_friction = I_cm * α, where I_cm is the moment of inertia about the center of mass and α is the angular acceleration. For a rolling cylinder, α = a_cm / R.

(a) Now, let's find the magnitude of the acceleration, a_cm, of the center of mass of the cylinder on the incline. Since F_friction_max = F_friction, we can equate the two expressions for the friction force:

μs * m * g * cos(θ) = I_cm * a_cm / R

Rearranging the equation, we can find the magnitude of the acceleration, a_cm:

a_cm = (μs * m * g * cos(θ))/ (I_cm / R)

where I_cm = m * R^2.

(a) The magnitude of the acceleration, a, of the center of mass of the cylinder on the incline is given by:

a = (μs * g * cos(θ))/ (1 + k^2)

where k = R/h.

(b) To find the minimum value for the coefficient of static friction, μs, we need to consider the limiting case where the cylinder is on the verge of slipping. This occurs when the static friction force is at its maximum value, F_friction_max. Setting F_friction_max equal to the maximum possible value of static friction, we have:

μs_min * m * g * cos(θ) = μs_max * m * g * cos(θ) = I_cm * a_cm / R

Simplifying, we find:

μs_min = (I_cm * a_cm) / (m * g * cos(θ))

(c) Finally, to find the magnitude of the velocity of the center of mass of the cylinder when it reaches the bottom of the incline, we can use the conservation of mechanical energy. The potential energy of the cylinder at the top of the incline is converted into kinetic energy at the bottom. The potential energy of the cylinder at the top is given by m * g * h, where h is the vertical distance dropped.

At the bottom of the incline, the total kinetic energy is equal to the translational kinetic energy of the center of mass and the rotational kinetic energy. The translational kinetic energy, KE_trans, is given by (1/2) * m * v_cm^2, where v_cm is the velocity of the center of mass of the cylinder. The rotational kinetic energy, KE_rot, is given by (1/2) * I_cm * (ω^2), where ω is the angular velocity.

Since the cylinder is rolling without slipping, the linear velocity and the angular velocity are related by v_cm = ω * R. Thus, the total kinetic energy at the bottom is given by:

KE_total = KE_trans + KE_rot = (1/2) * m * v_cm^2 + (1/2) * I_cm * (v_cm / R)^2

Setting the total kinetic energy equal to the potential energy at the top, we have:

m * g * h = (1/2) * m * v_cm^2 + (1/2) * I_cm * (v_cm / R)^2

Simplifying the equation and solving for v_cm, we find:

v_cm = sqrt((2 * g * h) / (1 + k^2))

where k = R/h.

(c) The magnitude of the velocity of the center of mass of the cylinder when it reaches the bottom of the incline is given by:

vf = sqrt((2 * g * h) / (1 + k^2))