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Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

A rolling member has both linear and angular degrees of freedom.

We know that: T = Iα , where [T = torque], [I = Mass moment of inertia] and
[α = angular acceleration]

The Newton’s second law, since friction provides angular inertia, we have:

T = Iα = μ mg cos θ r = Iα ,

a
We know that α in terms of its linear accelerations is
r
a
Therefore μ mg cos θ r = I - - - - (2.46)
r

Equation (2.46) gives the friction torque about the CM.

Simplifying equation (2.46), we get:

μ mg cos θ r 2
2
μ mg cos θ r = I a , or a = - - - - (2.47)
I
Acceleration (a) can be found by balancing the horizontal forces:

ma + μ mg cosθ = mg sinθ or a = g sin θ − μ g cos θ

Substituting the value of ‘a’ in equation (2.47), we get:

μ mg cos θ r 2 μ mg cos θ r 2
g sin θ − μ g cos θ = = g sin θ = + μ g cos θ
I I

μ mg cos θ r 2 mr 2
g sin θ = μg + cos θ = g sin θ = μg cos θ + 1 - - - - (2.48)
I I

Simplifying equation (2.48), we get minimum friction coefficient required to prevent the cylinder
from slipping.

tanθ
μ = mr 2
+ 1
I

It is to be noted that, the friction provides angular inertia for the cylinder. Therefore, the friction
torque must be greater than or equal to (Iα).

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
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