Motion of a Cylinder Rolling on a Inclined Plane

1. Concept Overview

A cylinder rolling down an incline can either roll without slipping (pure rolling) or slip (with kinetic friction). For rolling without slipping, there is a kinematic constraint:

[ v = \omega R ]

where [ v ] is the translational speed of the cylinder’s center, [ \omega ] its angular speed, and [ R ] its radius.

Two standard analysis approaches:

1. Newton’s second law + torque (dynamics): use forces and torques to get linear acceleration.
2. Energy method: Use conservation of mechanical energy (if no dissipative work) to relate translational and rotational energy.

2. Key Features

• The cylinder rolls without slipping down an inclined plane making an angle [\theta] with the horizontal.
• The frictional force prevents slipping but does no work since there is no relative motion at the point of contact.
• The motion involves both translational and rotational kinetic energy.
• The acceleration of the center of mass depends on the moment of inertia of the cylinder.

3. Derivation

Let the following be given:

• Mass of the cylinder = [M]
• Radius = [R]
• Angle of incline = [\theta]
• Acceleration of the center of mass = [a]
• Angular acceleration = [\alpha]

Since the cylinder rolls without slipping:
[
a = R \alpha
]

The forces acting on the cylinder are:

1. Component of weight along incline = [M g \sin \theta]
2. Frictional force (up the incline) = [f]

Applying Newton’s second law along the incline:
[
M g \sin \theta – f = M a \quad \text{…(1)}
]

About the center of mass, torque is:
[
f R = I \alpha
]

For a solid cylinder,
[
I = \dfrac{1}{2} M R^2
]

Substituting [\alpha = \dfrac{a}{R}]:
[
f R = \dfrac{1}{2} M R^2 \dfrac{a}{R}
]
[
f = \dfrac{1}{2} M a \quad \text{…(2)}
]

Substitute (2) in (1):
[
M g \sin \theta – \dfrac{1}{2} M a = M a
]
[
a = \dfrac{2}{3} g \sin \theta
]

Thus, the acceleration of the rolling cylinder is:

[
\boxed{a = \dfrac{2}{3} g \sin \theta}
]

The frictional force can be found from (2):

[
f = \dfrac{1}{3} M g \sin \theta
]

4. Energy Relation

For rolling without slipping, the total kinetic energy is:
[
K = \dfrac{1}{2} M v^2 + \dfrac{1}{2} I \omega^2
]
Substitute [I = \dfrac{1}{2} M R^2] and [\omega = \dfrac{v}{R}]:
[
K = \dfrac{1}{2} M v^2 + \dfrac{1}{4} M v^2 = \dfrac{3}{4} M v^2
]
So, the total kinetic energy is [\dfrac{3}{4} M v^2].

5. Important Formulas to Remember

6. Conceptual Questions

7. FAQ / Common Misconceptions

8. Practice Questions (with Solutions)

Q1. A solid cylinder rolls without slipping down a 30° incline. Find its acceleration.
Solution:
[
a = \dfrac{2}{3} g \sin 30° = \dfrac{2}{3} \times 9.8 \times \dfrac{1}{2} = 3.27 , \text{m/s}^2
]

Q2. What is the ratio of translational to rotational kinetic energy for a rolling cylinder?
Solution:
[\dfrac{K_{\text{trans}}}{K_{\text{rot}}}] [= \dfrac{\frac{1}{2} M v^2}{\frac{1}{4} M v^2} = 2:1]

Q3. A solid sphere and a solid cylinder roll down the same incline. Which reaches the bottom first?
Solution:
The sphere has a smaller moment of inertia relative to [M R^2], so it has greater acceleration and reaches first.

Q4. Find the frictional force on a 5 kg solid cylinder rolling down a 45° incline.
Solution:
[f = \dfrac{1}{3} M g \sin 45°] [= \dfrac{1}{3} (5)(9.8)(0.707) \approx 11.5 \text{N}]

Q5. Calculate the time taken by a cylinder to roll 2 m down a 30° incline from rest.
Solution:
[
a = \dfrac{2}{3} g \sin 30° = 3.27 , \text{m/s}^2
]
Using [s = \dfrac{1}{2} a t^2]:
[
t = \sqrt{\dfrac{2s}{a}} = \sqrt{\dfrac{4}{3.27}} = 1.10 , \text{s}
]