# Motion of an object on the outside of a cylinder

This question considers the motion of an object of mass m sliding on the outside of a cylinder of radius R whose axis is horizontal. The motion occurs in the vertical plane, and the surface of the cylinder is rough — the coefficient of sliding friction is μ'. The diagram below shows the position of the object when it is at an angle θ from the upward vertical through the axis of the cylinder O. Model the object as a particle.

Please see the attached MS Word document for the diagram and the followup questions.

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Please see the attached MS Word document for the full solution to the questions posed. Attached are formulas and diagrams to help clarify.

QUESTION

This question considers the motion of an object of mass m sliding on the outside of a cylinder of radius R whose axis is horizontal. The motion occurs in the vertical plane, and the surface of the cylinder is rough — the coefficient of sliding friction is μ'. The diagram below shows the position of the object when it is at an angle θ from the upward vertical through the axis of the cylinder O. Model the object as a particle.

1. Please draw a force diagram for the forces acting on the particle, defining any forces that you introduce.

2. Why is it better to use plane polar unit vectors. Please include them on the force diagram. How do I express the acceleration of the particle in terms of these unit vectors?

3. How do I express each of the forces acting on the particle as a vector in terms of the plane polar unit vectors?

4. How would I write down a differential equation of motion for the particle, and how would I then obtain two scalar differential equations?

5. If μ < tan θ, where μ is the coefficient of static friction how would I show that the particle will slide down the outside of the cylinder?

6. Why it is not possible to use conservation of energy for this system?

7. How would I use the friction condition to show that by eliminating the unknown magnitudes of forces in the two equations in part (d), this results in the equation

8. Use the substitution

How would I transform the differential equation in part 7 into

9. How would I express a, b and c in terms of μ' and g.

SOLUTION

1. The actual forces acting on the particle are:

- The weight G = mg , acting on the vertical direction downward;

- Normal reaction N to the cylinder surface;

- Friction force T, acting tangentially in the opposite direction to the movement

The weight G decomposes upon the polar vectors (tangential) and (normal) as follows:

( 1)

where the magnitudes of G and G are given by

( 2)

The actual forces are represented with red in the above figure; the "blue" forces are just the components of the weight G on the polar directions (they are not additional forces).

2. It is better to use the polar unit vectors (tangential) and (normal) because all the forces which determine the dynamic equilibrium are oriented upon these directions. This is why we have projected G force on the intrinsic axes of and , the other ones (respectively N and T) are already oriented upon and directions.

The acceleration of the particle occurs tangentially, that is upon direction, and it is determined by the difference between G and T, according to the 2-nd Newton's law:

( 3)

where m = mass of the particle, R = cylinder radius, g = gravitational acceleration

Therefore:

( 4)

Remark: In order for the particle to start sliding, it is necessary that

( 5)

3. By using Coulomb's law when the particle slides, the friction force can be expressed as:

( 6)

The normal force N is the difference between the component G and the centrifugal force:

( 7)

so that

( 8)

Replacing in (5), the friction force will be:

( 9)

In vector form, the expressions of the above forces are:

( 10)

(the "minus" sign from T is because this force is oriented in opposite direction about )

4. From (3), one derives the equation:

( 11)

Thus, the expressions (9) and (11) are the scalar differential equations which describe the movement of the particle.

5. Replacing T in (5) with (9) and considering the particle at rest ( ), the condition for the particle to slide becomes:

( 12)

Here, ' is the coefficient of dynamic friction, which is smaller than = coefficient of static friction. Since > ' and it is stated that < tan, it follows than ' < tan and the condition (12) of sliding is satisfied, so that the particle will slide down.

6. Whenever the friction occurs, the law of conservation of energy cannot be applied.

This is because some energy is lost irreversibly by dissipation as heat due to the friction.

7. Substituting (9) in (11), one yields:

( 13)

8. If one denotes

( 14)

it follows that

( 15)

With this, the differential equation (13) becomes:

( 16)

9. If the differential equation (16) is written as

( 17)

then, by identification, one finds:

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