Question:

A solid uniform cylinder of mass M and radius R rolls without slipping down an inclined plane starting from rest. The incline's angle of elevation with respect to the horizontal is θ. The moment of inertia of the cylinder about its central axis is given by I = (1/2)MR^2, where M is the mass of the cylinder and R is its radius.

(a) Determine the acceleration of the cylinder as it rolls down the incline in terms of M, R, and θ.

(b) Calculate the minimum angle of incline (θ_min) required for the cylinder to start rolling down the incline without sliding.

(c) If the cylinder starts from rest at an angle θ = θ_min and rolls down the incline, calculate the speed of the cylinder at the bottom of the incline in terms of M, R, and g, where g is the acceleration due to gravity.

Answer:

(a) To find the acceleration of the cylinder as it rolls down the incline, we can analyze the forces acting on it. The gravitational force pulls the cylinder down the incline and can be resolved into two components: mg sin(θ) parallel to the incline and mg cos(θ) perpendicular to the incline. The normal force exerted by the incline cancels out the perpendicular component of gravity. Additionally, there is a frictional force opposing the motion of the cylinder.

Since the cylinder is rolling without slipping, the frictional force can be expressed as f = μN, where μ is the coefficient of friction and N is the normal force. The normal force can be determined as N = mg cos(θ).

Using Newton's second law, we can write the equation of motion for the cylinder:

ΣF = ma

Here, the net force acting on the cylinder is the component of gravitational force parallel to the incline minus the frictional force:

mg sin(θ) - μN = ma

Substituting for N, we get:

mg sin(θ) - μ(mg cos(θ)) = ma

Simplifying,

a = g (sin(θ) - μ cos(θ))

Therefore, the acceleration of the cylinder as it rolls down the incline is a = g (sin(θ) - μ cos(θ)).

(b) The condition for the cylinder to start rolling down the incline without sliding is when the static frictional force reaches its maximum value (before sliding begins). The maximum value of static friction is given by fs = μsN, where μs is the coefficient of static friction.

When the cylinder is at the point of impending sliding, the static frictional force must be equal to the gravitational force component parallel to the incline:

fs = mg sin(θ)

Substituting μsN for fs,

μsN = mg sin(θ)

Since N = mg cos(θ),

μs(mg cos(θ)) = mg sin(θ)

μs cos(θ) = sin(θ)

Dividing both sides by cos(θ),

μs = tan(θ)

Therefore, the minimum angle of incline required for the cylinder to start rolling down the incline without sliding is θ_min = tan⁻¹(μs).

(c) To determine the speed of the cylinder at the bottom of the incline, we can use the conservation of energy principle. The potential energy of the cylinder at the starting point is converted into both translational kinetic energy and rotational kinetic energy at the bottom.

The initial potential energy of the cylinder is given by U = mgh, where h is the height of the incline. At the bottom of the incline, all of the potential energy is converted into translational and rotational kinetic energy.

Translational kinetic energy (Kt) is given by Kt = (1/2)mv², where v is the velocity of the center of mass. Rotational kinetic energy (Kr) is given by Kr = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity.

At the bottom, the entire potential energy has been transformed, so:

U = Kt + Kr

Substituting the equations for the respective kinetic energies and the given moment of inertia:

mgh = (1/2)mv² + (1/2)(1/2)MR²ω²

The velocity of the center of mass can be related to the angular velocity as v = Rω, since the cylinder is rolling without slipping.

h = (1/2) v²/R + (1/4)R²ω²

Since the cylinder is rolling without slipping, at the bottom of the incline, the linear speed of the center of mass is related to the angular velocity as v = Rω.

h = (1/2)(Rω)²/R + (1/4)R²ω²

h = (1/2)R²ω² + (1/4)R²ω²

h = (3/4)R²ω²

Rearranging the equation,

ω² = (4h)/(3R²)

Using the relationship ω = v/R,

(v/R)² = (4h)/(3R²)

Simplifying,

v² = (4hR²)/3

At the bottom of the incline, h = 0, so:

v² = (4×0×(R²))/3

v = 0

Therefore, at the bottom of the incline, the speed of the cylinder is v = 0.