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

For a hoop sliding down the inclined plane, the acceleration is given by:

g sin θ
a =
1 + I
mr 2
As for sliding, there is no rotational inertia, the mass moment of inertia (I) is zero. Therefore, the
acceleration for a hoop sliding down an inclined plane is given by:

g sinθ
a = = g sin θ
1 + 0

considering the conservation of energy, as there is no loss of energy.

Potential Energy = Transational Kinetic Energy + Rotational Kinetic Energy

1 1 1 1 v 2 1
2
2
2
2
mgH = mv + Iω = mv + ∗ 0 ∗ = mv
2 2 2 2 r 2 2
Further simplifying the above expression, we get:
1
v = gH = v = 2gH
2
2
To compute the time of travel, let us use the expression [v = U+at], where, the initial velocity ‘U’ is
zero.

Substituting the value of ‘v’ and ‘a’ in the expression [v = U+at], we get:

1 2H
2gH = 0 + g sinθ ∗ t = t = 2gH ∗ = t = - - - - (2.53)
2
g sinθ g sin θ

Dividing equation (2.53) by equation (2.52), gives us the ratio of time taken by the hoop slide and
roll down the inclined plane. Therefore, we have:

2 H
2
t sliding g sin θ 2 H 2
t = 2H = 2 H = 1 - - - - (2.54)

rolling 2
g sin θ
It can be seen from equation (2.54) that, the sliding time is [ 2 = 1.414] times more than that of
rolling time.

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 75 age 75
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