A Classical Approach of Newtonian Mechanics

8 ROTATIONAL MOTION 8.11 Combined translational and rotational motion

×

(1/2) M b^2 , and so
̇vsolid =

2

g sin θ. (8.104)

3

For the case of the hollow cylinder, the moment of inertia is I = M b^2 (i.e., the
same as that of a ring with a similar mass, radius, and axis of rotation), and so

̇vhollow =

1

g sin θ. (8.105)

2

It is clear that the solid cylinder reaches the bottom of the slope before the hollow

one (since it possesses the greater acceleration). Note that the accelerations of
the two cylinders are independent of their sizes or masses. This suggests that a

solid cylinder will always roll down a frictional incline faster than a hollow one,

irrespective of their relative dimensions (assuming that they both roll without
slipping). In fact, Eq. (8.99) suggests that whenever two different objects roll

(without slipping) down the same slope, then the most compact object—i.e., the

object with the smallest I/M b^2 ratio—always wins the race.

Worked example 8.1: Balancing tires

Question: A tire placed on a balancing machine in a service station starts from

rest and turns through 5.3 revolutions in 2.3 s before reaching its final angular
speed. What is the angular acceleration of the tire (assuming that this quantity
remains constant)? What is the final angular speed of the tire?

Answer: The tire turns through φ = 5.3 2 π = 33.30 rad. in t = 2.3 s. The
relationship between φ and t for the case of rotational motion, starting from rest,

with uniform angular acceleration α is

φ =

1

α t^2.

2

Hence,
2 φ
α =
t^2

=^

2 × 33.30

2.3^2

= 12.59 rad./s^2.